Air, Water and Solubility Practice Questions — JAMB UTME — Page 3

Question 41Question →

The solubility of a salt $Y$ (molar mass = $100\text{ g/mol}$ ) in water is $3.5\text{ mol/dm}^3$ at $70^\circ\text{C}$ and $1.5\text{ mol/dm}^3$ at $25^\circ\text{C}$ . What mass of salt $Y$ will crystallize out when $500\text{ cm}^3$ of its saturated solution is cooled from $70^\circ\text{C}$ to $25^\circ\text{C}$ ?

100.0\text{ g}

175.0\text{ g}

200.0\text{ g}

1.0\text{ g}

Show answer & explanation

Answer: $100.0\text{ g}$

Answer

The mass of salt

Y

that crystallizes out of solution is

100.0\text{ g}

.

Cooling a saturated solution reduces its capacity to retain solute. The difference in concentration between

70^\circ\text{C}

(

3.5\text{ mol/dm}^3

) and

25^\circ\text{C}

(

1.5\text{ mol/dm}^3

) is

2.0\text{ mol/dm}^3

. Scaling this concentration difference for

500\text{ cm}^3

(

0.5\text{ dm}^3

) gives

1.0\text{ mole}

of salt

Y

. Converting moles to mass yields

1.0\text{ mole} \times 100\text{ g/mol} = 100.0\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in solubility between

70^\circ\text{C}

and

25^\circ\text{C}

in

\text{mol/dm}^3

.

\Delta S = 3.5\text{ mol/dm}^3 - 1.5\text{ mol/dm}^3 = 2.0\text{ mol/dm}^3

Solubility decreases as temperature drops, causing solute to precipitate.

2

Convert solution volume from

\text{cm}^3

to

\text{dm}^3

and determine the moles of solute precipitated.

V = \frac{500\text{ cm}^3}{1000} = 0.5\text{ dm}^3

;

\text{Moles precipitated} = 2.0\text{ mol/dm}^3 \times 0.5\text{ dm}^3 = 1.0\text{ mole}

Solubility is given per cubic decimeter, so volume scaling is necessary.

3

Convert the precipitated moles into mass in grams using molar mass.

\text{Mass} = 1.0\text{ mole} \times 100\text{ g/mol} = 100.0\text{ g}

Mass equals number of moles multiplied by molar mass.

Key Concept

Crystallization calculations based on temperature-dependent solubility curves

Question 42Question →

Match each type of solution state with its corresponding characteristic physical state and behavior at a given temperature.

Click a left item, then click its matching right item

Show answer & explanation

Answer

An unsaturated solution matches with holding less solute than the maximum limit and being able to dissolve more solute. A saturated solution matches with holding the maximum solute in dynamic equilibrium with undissolved solute. A supersaturated solution matches with holding excess dissolved solute beyond normal solubility and rapidly crystallizing upon seeding.

An unsaturated solution can dissolve more solute; a saturated solution maintains dynamic equilibrium with maximum dissolved solute; a supersaturated solution holds solute beyond normal solubility limits and crystallizes when seeded.

Step-by-Step Solution

1

Identify the characteristic of an unsaturated solution.

It has not reached maximum capacity, so adding solute leads to further dissolution.

Solvent molecules are available to interact with and dissolve additional solute particles at that temperature.

2

Identify the characteristic of a saturated solution.

It exists in dynamic equilibrium with undissolved solid.

At saturation, the solvent holds the maximum possible concentration of solute at that specific temperature.

3

Identify the characteristic of a supersaturated solution.

It contains dissolved solute in excess of the saturation concentration and is metastable/unstable.

Disturbing the solution with a seed crystal initiates rapid crystallization to relieve the unstable excess concentration.

Key Concept

Saturation States of Solutions

Question 43Question →

The solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water is $3.5\text{ mol dm}^{-3}$ at $70^\circ\text{C}$ and $1.2\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . What mass of $\text{KNO}_3$ will crystallize out of solution when $200\text{ cm}^3$ of a saturated solution at $70^\circ\text{C}$ is cooled to $25^\circ\text{C}$ ? $[\text{K} = 39, \text{N} = 14, \text{O} = 16]$

46.46\text{ g}

0.46\text{ g}

70.70\text{ g}

24.24\text{ g}

Show answer & explanation

Answer: $46.46\text{ g}$

Answer

The mass of potassium trioxonitrate(V) that crystallizes out is

46.46\text{ g}

.

The net solubility decrease when cooling from

70^\circ\text{C}

to

25^\circ\text{C}

is

2.3\text{ mol dm}^{-3}

. In

200\text{ cm}^3

(

0.2\text{ dm}^3

) of saturated solution,

0.46\text{ mol}

of

\text{KNO}_3

precipitates out. Multiplying

0.46\text{ mol}

by the molar mass of

\text{KNO}_3

(

101\text{ g mol}^{-1}

) yields

46.46\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of potassium trioxonitrate(V),

\text{KNO}_3

.

\text{Molar mass} = 39 + 14 + (3 \times 16) = 101\text{ g mol}^{-1}

.

Molar mass is needed to convert molar concentration into mass.

2

Find the change in solubility per

\text{dm}^3

upon cooling from

70^\circ\text{C}

to

25^\circ\text{C}

.

\Delta S = 3.5\text{ mol dm}^{-3} - 1.2\text{ mol dm}^{-3} = 2.3\text{ mol dm}^{-3}

.

The crystallization amount depends on the difference between initial and final solubilities.

3

Determine the amount of solute precipitated in

200\text{ cm}^3

(

0.2\text{ dm}^3

) of solution.

\text{Moles} = 2.3\text{ mol dm}^{-3} \times 0.2\text{ dm}^3 = 0.46\text{ mol}

.

Solubility values are given per

\text{dm}^3

, so they must be scaled to the given volume of

200\text{ cm}^3

.

4

Convert the precipitated moles into mass in grams.

\text{Mass} = 0.46\text{ mol} \times 101\text{ g mol}^{-1} = 46.46\text{ g}

.

Multiplying the number of moles by molar mass gives the required mass in grams.

Key Concept

Solubility Curves and Temperature Effects

Question 44Question →

Match each solubility phenomenon or term on the left with its correct thermodynamic or practical description on the right. Which pair correctly connects each term to its appropriate description?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fractional crystallization pairs with separation technique using solubility differences; Endothermic dissolution pairs with heat absorption increasing solubility with temperature; Exothermic dissolution pairs with heat evolution decreasing solubility with temperature; Supersaturated solution pairs with holding excess dissolved solute beyond equilibrium limit.

Each term directly aligns with its fundamental thermodynamic property or chemical application. Fractional crystallization isolates salts using temperature-dependent solubility variations; endothermic dissolution absorbs heat so solubility increases with temperature; exothermic dissolution gives off heat so solubility decreases with temperature; and a supersaturated solution holds excess dissolved solute beyond standard equilibrium capacity.

Step-by-Step Solution

1

Identify the separation technique based on solubility curve variance.

Fractional crystallization separates solutes based on their distinct solubility curves across temperatures.

Differences in solubility curves allow one component to crystallize out of solution before another when cooled.

2

Analyze how enthalpy changes govern temperature dependence on solubility.

Endothermic processes absorb heat leading to increased solubility with temperature rise, whereas exothermic processes release heat leading to decreased solubility with temperature rise.

According to Le Chatelier's principle, adding heat favors the endothermic direction of a solution equilibrium.

3

Determine the saturation state definition for excess solute concentration.

A supersaturated solution contains a higher concentration of dissolved solute than a saturated solution at the specified temperature.

It represents a metastable condition created by careful cooling without crystallization.

Key Concept

Temperature effects on solubility, enthalpy of solution, and fractional crystallization

Question 45Question →

The table below shows the solubility of anhydrous copper(II) tetraoxosulfate(VI), $\text{CuSO}_4$ , in water at different temperatures:

Temperature ( $^\circ\text{C}$ )	Solubility ( $\text{g}$ of $\text{CuSO}_4$ per $100\text{ g}$ of $\text{H}_2\text{O}$ )
$20$	$20.0$
$40$	$29.0$
$80$	$55.0$

A saturated solution of copper(II) tetraoxosulfate(VI) in $200.0\text{ g}$ of water at $80^\circ\text{C}$ is cooled to $20^\circ\text{C}$ . Given that the molar mass of $\text{CuSO}_4 = 160\text{ g mol}^{-1}$ and $\text{H}_2\text{O} = 18\text{ g mol}^{-1}$ , what is the exact mass of hydrated copper(II) tetraoxosulfate(VI) pentahydrate crystals, $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$ , that will deposit from the solution?

123.24\text{ g}

109.38\text{ g}

70.00\text{ g}

44.80\text{ g}

Show answer & explanation

Answer: $123.24\text{ g}$

Answer

The mass of copper(II) tetraoxosulfate(VI) pentahydrate crystals deposited is

123.24\text{ g}

.

When a hydrated salt crystallizes, it removes both solute and water of crystallization from the saturated solution. Taking into account that

m

grams of

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

contains

0.64m

grams of

\text{CuSO}_4

and

0.36m

grams of

\text{H}_2\text{O}

, setting up the solubility ratio at

20^\circ\text{C}

as

(110.0 - 0.64m) / (200.0 - 0.36m) = 0.20

gives

m = 123.24\text{ g}

.

Step-by-Step Solution

1

Calculate the initial mass of dissolved anhydrous

\text{CuSO}_4

at

80^\circ\text{C}

.

In

200.0\text{ g}

of water, mass of dissolved

\text{CuSO}_4 = 2 \times 55.0\text{ g} = 110.0\text{ g}

.

Solubility at

80^\circ\text{C}

is

55.0\text{ g}

per

100\text{ g}

of water.

2

Determine the molar masses of the anhydrous salt, water, and hydrate.

Molar mass of

\text{CuSO}_4 = 160\text{ g mol}^{-1}

;

\text{H}_2\text{O} = 18\text{ g mol}^{-1}

;

\text{CuSO}_4\cdot 5\text{H}_2\text{O} = 160 + 5(18) = 250\text{ g mol}^{-1}

.

Needed to establish mass fractions of solute and solvent in the crystals.

3

Express the mass fractions of anhydrous salt and water in

m

grams of

\text{CuSO}_4\cdot 5\text{H}_2\text{O}

crystals.

Anhydrous

\text{CuSO}_4

fraction

= \frac{160}{250}m = 0.64m

; Water fraction

= \frac{90}{250}m = 0.36m

.

As crystals form, they take away both anhydrous salt and water from the solution.

4

Set up the solubility saturation equation at

20^\circ\text{C}

.

\frac{110.0 - 0.64m}{200.0 - 0.36m} = \frac{20.0}{100.0} = 0.20

.

At

20^\circ\text{C}

, the remaining solution must remain saturated.

5

Solve for

m

.

110.0 - 0.64m = 0.20(200.0 - 0.36m) \implies 110.0 - 0.64m = 40.0 - 0.072m \implies 70.0 = 0.568m \implies m = 123.24\text{ g}

.

Isolating

m

gives the exact mass of hydrated crystals deposited.

Key Concept

Crystallization of Hydrated Salts from Saturated Solutions

Question 46Question →

In a municipal water treatment system, the effluent water following alum coagulation and filtration is found to be slightly acidic and retains a persistent earthy odor. Which pair of chemical substances should be added to neutralize the acidity and eliminate the odor, respectively?

Slaked lime and activated carbon

Calcium tetraoxosulfate(VI) and chlorine gas

Washing soda and potash alum

Concentrated tetraoxosulfate(VI) acid and bleaching powder

Show answer & explanation

Answer: Slaked lime and activated carbon

Answer

Slaked lime (

\text{Ca(OH)}_2

) or soda ash is added to neutralize acidity (raising pH), while activated carbon (charcoal) is used to remove unpleasant odors and tastes via adsorption.

Slaked lime (

\text{Ca(OH)}_2

) acts as a base to neutralize hydrogen ions, correcting acidity, while activated carbon effectively adsorbs volatile organic compounds that cause unpleasant tastes and odors in water.

Step-by-Step Solution

1

Identify the cause of acidity and the substance required to raise pH.

Coagulation with alum leaves acidic residues due to hydrolysis. An alkali such as slaked lime,

\text{Ca(OH)}_2

, neutralizes acidity.

Acids are neutralized by adding basic compounds to adjust the water pH to a safe drinking level.

2

Identify the process and agent required for odor removal.

Activated carbon (charcoal) possesses a high surface area that adsorbs dissolved organic volatile impurities responsible for odor.

Physical adsorption on activated carbon removes taste and odor without introducing additional chemical pollutants.

Key Concept

Chemical roles in water purification: pH adjustment using slaked lime/soda ash and taste/odor removal using activated carbon.

Estimated Time:1m 0s

Question 47Question →

Match each chemical compound on the left with its characteristic solubility trend in water as temperature increases on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium trioxonitrate(V) matches with steep increase in solubility; Sodium chloride matches with nearly constant solubility; Calcium tetraoxosulfate(VI) matches with decreasing solubility as temperature rises.

Potassium trioxonitrate(V) shows a steep increase in solubility with rising temperature due to its endothermic nature. Sodium chloride exhibits minimal temperature sensitivity, keeping its curve nearly flat. Calcium tetraoxosulfate(VI) exhibits retrograde solubility, decreasing as temperature increases because its dissolution is exothermic.

Step-by-Step Solution

1

Identify the thermodynamic enthalpy change associated with dissolving each salt in water.

Potassium trioxonitrate(V) dissolution is endothermic, sodium chloride dissolution has a near-zero enthalpy change, and calcium tetraoxosulfate(VI) dissolution is exothermic.

Le Chatelier's principle determines how temperature affects solubility equilibria based on whether heat is absorbed or released.

2

Relate enthalpy of solution to the slope of the solubility curve.

Potassium trioxonitrate(V) has a steep positive curve, sodium chloride has a nearly horizontal curve, and calcium tetraoxosulfate(VI) has a negative curve.

Endothermic dissolution shifts right with heat (increasing solubility), whereas exothermic dissolution shifts left with heat (decreasing solubility).

Key Concept

Solubility curves represent how solute solubility varies with temperature based on whether the dissolution process is endothermic or exothermic.

Question 48Question →

During the municipal treatment of river water containing dissolved iron(II) salts and unpleasant odors caused by dissolved hydrogen sulfide gas, raw water is initially sprayed into the air during an aeration process. What is the primary chemical purpose of this aeration step prior to coagulation?

To oxidize soluble iron(II) compounds into insoluble iron(III) hydroxide while expelling volatile dissolved gases

To destroy pathogenic bacteria and harmful microorganisms present in the raw water supply

To precipitate calcium and magnesium hydrogen trioxocarbonates(IV) causing temporary water hardness

To cause fine colloidal clay suspensions to aggregate into larger settleable flocs

Show answer & explanation

Answer: To oxidize soluble iron(II) compounds into insoluble iron(III) hydroxide while expelling volatile dissolved gases

Answer

Aeration oxidizes soluble iron(II) compounds into insoluble iron(III) hydroxide and expels volatile dissolved gases from the water.

Aeration introduces atmospheric oxygen into raw water, which oxidizes soluble iron(II) compounds into an insoluble iron(III) hydroxide precipitate (

Fe(OH)_3

). Additionally, the splashing action strips out volatile gases responsible for bad tastes and odors, such as hydrogen sulfide (

H_2S

).

Step-by-Step Solution

1

Identify the chemical changes occurring when air contacts raw water during aeration.

Dissolved atmospheric oxygen reacts with dissolved

Fe^{2+}

ions, while volatile gases like

H_2S

escape into the atmosphere.

Aeration increases dissolved oxygen concentration and promotes liquid-gas equilibrium exchange.

2

Determine the resulting chemical products formed.

Soluble iron(II) ions form insoluble iron(III) hydroxide,

Fe(OH)_3

, precipitate.

Insoluble precipitates formed during aeration can subsequently be settled and removed during sedimentation and filtration.

Key Concept

Aeration in Municipal Water Purification

Estimated Time:1m 0s

Question 49Question →

Solve the following solubility and crystallization problem. What is the mass of the salt that crystallizes out of the solution upon cooling?

Fill in the blanks below

A saturated solution of a divalent metal salt

\text{MX}_2

(molar mass =

120\text{ g/mol}

) has a total mass of

120.0\text{ g}

at

60^\circ\text{C}

. The solubility of

\text{MX}_2

is

5.0\text{ mol/dm}^3

at

60^\circ\text{C}

and

2.0\text{ mol/dm}^3

at

25^\circ\text{C}

. Assuming the density of water is

1.00\text{ g/cm}^3

, the mass of

\text{MX}_2

deposited when the solution is cooled from

60^\circ\text{C}

to

25^\circ\text{C}

is g.

Show answer & explanation

Answer

The mass of

\text{MX}_2

deposited on cooling is

27.0\text{ g}

.

At

60^\circ\text{C}

, a solubility of

5.0\text{ mol/dm}^3

corresponds to

5.0 \times 120 = 600\text{ g}

of

\text{MX}_2

per

1000\text{ g}

of water. Thus,

1600\text{ g}

of saturated solution contains

600\text{ g}

solute and

1000\text{ g}

water. Proportionally,

120.0\text{ g}

of saturated solution contains

45.0\text{ g}

of

\text{MX}_2

dissolved in

75.0\text{ g}

of water. At

25^\circ\text{C}

, the solubility decreases to

2.0\text{ mol/dm}^3

, which equals

2.0 \times 120 = 240\text{ g}

of

\text{MX}_2

per

1000\text{ g}

of water. In

75.0\text{ g}

of water, the maximum mass of solute that remains dissolved is

75.0 \times (240 / 1000) = 18.0\text{ g}

. The mass of solid

\text{MX}_2

that crystallizes out is

45.0\text{ g} - 18.0\text{ g} = 27.0\text{ g}

.

Step-by-Step Solution

1

Calculate the mass of solute per

1000\text{ g}

of water at

60^\circ\text{C}

.

Mass of

\text{MX}_2

per

1000\text{ g}

water =

5.0\text{ mol/dm}^3 \times 120\text{ g/mol} = 600.0\text{ g}

.

Converting solubility from

\text{mol/dm}^3

to

\text{g/dm}^3

(or grams per

1000\text{ g}

of water, given water density is

1.00\text{ g/cm}^3

).

2

Determine the composition of the

120.0\text{ g}

saturated solution at

60^\circ\text{C}

.

Total mass of saturated solution containing

1000\text{ g}

water =

1000\text{ g} + 600\text{ g} = 1600.0\text{ g}

. Mass of water in

120.0\text{ g}

solution =

120.0\text{ g} \times \frac{1000}{1600} = 75.0\text{ g}

. Mass of

\text{MX}_2

dissolved =

120.0\text{ g} - 75.0\text{ g} = 45.0\text{ g}

.

To find how much solute and solvent are actually present in the given portion of solution.

3

Calculate the mass of solute that remains dissolved in

75.0\text{ g}

of water at

25^\circ\text{C}

.

At

25^\circ\text{C}

, mass of solute per

1000\text{ g}

water =

2.0\text{ mol/dm}^3 \times 120\text{ g/mol} = 240.0\text{ g}

. Mass dissolved in

75.0\text{ g}

water =

75.0\text{ g} \times \frac{240.0}{1000} = 18.0\text{ g}

.

Determining the maximum amount of solute that

75.0\text{ g}

of water can hold at the lower temperature.

4

Calculate the mass of

\text{MX}_2

deposited upon cooling.

Mass deposited =

45.0\text{ g} - 18.0\text{ g} = 27.0\text{ g}

.

The difference between the initial mass dissolved at

60^\circ\text{C}

and the remaining mass dissolved at

25^\circ\text{C}

represents the crystallized solid.

Key Concept

Quantitative solubility calculations involving crystallization from saturated solutions

Question 50Question →

The solubility of a salt $X$ is $0.80\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $0.30\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . Calculate the mass of salt $X$ (in grams) that will crystallize out of solution when $1.0\text{ dm}^3$ of its saturated solution is cooled from $60^\circ\text{C}$ to $20^\circ\text{C}$ . (Molar mass of salt $X = 100\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 50

Answer

The mass of salt

X

that crystallizes out of the solution is

50\text{ g}

.

At

60^\circ\text{C}

,

1.0\text{ dm}^3

of saturated solution contains

0.80\text{ mol}

of salt

X

. When cooled to

20^\circ\text{C}

, the solution can only hold

0.30\text{ mol}

. The excess amount that crystallizes out is

0.80\text{ mol} - 0.30\text{ mol} = 0.50\text{ mol}

. Converting this amount to mass yields

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in molar solubility between the two temperatures

0.80\text{ mol dm}^{-3} - 0.30\text{ mol dm}^{-3} = 0.50\text{ mol dm}^{-3}

This difference represents the amount of solute in moles per cubic decimeter that can no longer remain dissolved when cooled to

20^\circ\text{C}

.

2

Convert the precipitated moles into mass in grams for

1.0\text{ dm}^3

of solution

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

Multiplying the precipitated amount in moles by the molar mass gives the total mass in grams that crystallizes out.

Key Concept

Solubility and crystallization calculations upon cooling

Question 51Question →

A laboratory technician prepares a mixture by thoroughly stirring gelatin in warm water. The resulting mixture passes completely through standard filter paper without leaving any solid residue, but when a concentrated beam of light passes through it in a darkened box, the path of the light beam becomes illuminated and visible. Which type of mixture is represented by this system?

A colloidal system

A true solution

A suspension

A coarse sediment

Show answer & explanation

Answer: A colloidal system

Answer

A colloidal system

Gelatin dispersed in warm water is a colloidal system. Colloidal particles range in diameter from

1\text{ nm}

to

100\text{ nm}

. This particle size is small enough to pass through standard filter paper pores, but large enough to scatter visible light rays (Tyndall effect).

Step-by-Step Solution

1

Analyze the filtration behavior described in the stem.

The mixture passes through standard filter paper without leaving a residue.

This eliminates suspensions, which have particle diameters exceeding

100\text{ nm}

and get trapped by filter paper pores.

2

Analyze the optical behavior (light scattering) described in the stem.

The mixture demonstrates the Tyndall effect by illuminating the light path.

True solutions have solute particles smaller than

1\text{ nm}

that cannot scatter visible light, whereas colloidal particles (

1\text{ nm}

to

100\text{ nm}

) effectively scatter light.

3

Synthesize the particle size characteristics to identify the mixture class.

The mixture is a colloidal system.

Exhibiting light scattering while passing through filter paper is the hallmark property distinguishing colloidal systems from true solutions and suspensions.

Key Concept

Distinguishing characteristics of true solutions, colloidal systems, and suspensions based on particle size and Tyndall effect.

Question 52Question →

The solubility of sodium trioxonitrate(V), $\text{NaNO}_3$ , in water is $4.5\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $2.0\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . If the molar mass of $\text{NaNO}_3$ is $85\text{ g mol}^{-1}$ , what mass of $\text{NaNO}_3$ in grams will crystallize out when $400\text{ cm}^3$ of a saturated solution of the salt at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

Show answer & explanation

Answer: 85

Answer

The mass of

\text{NaNO}_3

that will crystallize out is

85\text{ g}

.

The difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

is

4.5 - 2.0 = 2.5\text{ mol dm}^{-3}

. In

400\text{ cm}^3

(

0.4\text{ dm}^3

) of solution, the amount of salt precipitated is

2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

. Multiplying by the molar mass (

85\text{ g mol}^{-1}

) gives

85\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

\Delta S = 4.5\text{ mol dm}^{-3} - 2.0\text{ mol dm}^{-3} = 2.5\text{ mol dm}^{-3}

Cooling causes the excess solute to precipitate out based on the difference in saturation concentration.

2

Convert solution volume to

\text{dm}^3

V = \frac{400\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.4\text{ dm}^3

Solubility is given per

\text{dm}^3

, so volume must be in

\text{dm}^3

.

3

Calculate the moles of solute precipitated

n = 2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

Multiplying the concentration difference by the solution volume yields total precipitated moles.

4

Convert moles to mass in grams

m = 1.0\text{ mol} \times 85\text{ g mol}^{-1} = 85\text{ g}

Mass is found by multiplying moles by molar mass.

Key Concept

Crystallization and solubility change with temperature

Question 53Question →

What is the solubility in $\text{mol/dm}^3$ of potassium trioxonitrate(V), $\text{KNO}_3$ , at $40^\circ\text{C}$ if $50.5\text{ g}$ of the salt dissolves completely in $250\text{ cm}^3$ of water to form a saturated solution? $[\text{K} = 39, \text{N} = 14, \text{O} = 16]$

0.50\text{ mol/dm}^3

2.00\text{ mol/dm}^3

5.05\text{ mol/dm}^3

202.00\text{ mol/dm}^3

Show answer & explanation

Answer: $2.00\text{ mol/dm}^3$

Answer

The solubility of potassium trioxonitrate(V) at

40^\circ\text{C}

is

2.00\text{ mol/dm}^3

.

The correct answer is

2.00\text{ mol/dm}^3

. First, determine the molar mass of

\text{KNO}_3

, which is

39 + 14 + (3 \times 16) = 101\text{ g/mol}

. Converting

50.5\text{ g}

of

\text{KNO}_3

into moles gives

0.50\text{ mol}

. Converting

250\text{ cm}^3

of water into

\text{dm}^3

gives

0.25\text{ dm}^3

. Dividing

0.50\text{ mol}

by

0.25\text{ dm}^3

yields a molar solubility of

2.00\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KNO}_3

\text{Molar mass of KNO}_3 = 39 + 14 + 3(16) = 101\text{ g/mol}

Molar mass is required to convert the given mass of salt into number of moles.

2

Convert the mass of

\text{KNO}_3

to moles

\text{Moles of KNO}_3 = \frac{50.5\text{ g}}{101\text{ g/mol}} = 0.50\text{ mol}

Solubility in

\text{mol/dm}^3

requires solute quantity to be in moles.

3

Convert the volume of water from

\text{cm}^3

to

\text{dm}^3

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3

Standard molarity and molar solubility use cubic decimeters (

\text{dm}^3

) as the volume unit.

4

Calculate molar solubility

\text{Solubility} = \frac{0.50\text{ mol}}{0.25\text{ dm}^3} = 2.00\text{ mol/dm}^3

Solubility in

\text{mol/dm}^3

is defined as moles of solute per cubic decimeter of saturated solution.

Key Concept

Solubility in

\text{mol/dm}^3

measures the maximum number of moles of solute that dissolve in

1\text{ dm}^3

of solvent at a specific temperature.

Question 54Question →

Calculate the solubility of the salt in the given solution and complete the statement below.

Fill in the blanks below

If

5.85\text{ g}

of sodium chloride (

\text{NaCl}

) is dissolved in water to prepare

250\text{ cm}^3

of a saturated solution at

25^\circ\text{C}

, the solubility of

\text{NaCl}

is

\text{mol/dm}^3

.

[\text{Na} = 23.0, \text{Cl} = 35.5]

Show answer & explanation

Answer

The solubility of sodium chloride at

25^\circ\text{C}

is

0.4\text{ mol/dm}^3

.

The solubility in

\text{mol/dm}^3

is obtained by dividing the moles of solute by the solution volume in

\text{dm}^3

. Here,

5.85\text{ g}

of

\text{NaCl}

corresponds to

0.1\text{ mol}

(

5.85 / 58.5

), and

250\text{ cm}^3

equals

0.25\text{ dm}^3

. Dividing

0.1\text{ mol}

by

0.25\text{ dm}^3

gives

0.4\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of sodium chloride (

\text{NaCl}

).

Molar mass

= 23.0 + 35.5 = 58.5\text{ g/mol}

.

The molar mass is required to convert mass in grams to amount in moles.

2

Determine the number of moles of

\text{NaCl}

present.

\text{Moles} = \frac{5.85\text{ g}}{58.5\text{ g/mol}} = 0.1\text{ mol}

.

Solubility in

\text{mol/dm}^3

requires the quantity of solute in moles.

3

Convert the volume of solution from

\text{cm}^3

to

\text{dm}^3

.

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3

.

Concentration units are per cubic decimeter (

\text{dm}^3

).

4

Calculate the molar concentration (solubility).

\text{Solubility} = \frac{0.1\text{ mol}}{0.25\text{ dm}^3} = 0.4\text{ mol/dm}^3

.

Solubility is calculated as moles of solute divided by volume of solution in

\text{dm}^3

.

Key Concept

Solubility and Concentration Determination

Estimated Time:45s

Question 55Question →

The solubility of a salt $Z$ (molar mass = $101.0\text{ g mol}^{-1}$ ) in water is $5.0\text{ mol dm}^{-3}$ at $80^\circ\text{C}$ and $202.0\text{ g dm}^{-3}$ at $25^\circ\text{C}$ . If $250\text{ cm}^3$ of a saturated solution of salt $Z$ is cooled from $80^\circ\text{C}$ to $25^\circ\text{C}$ , what mass of salt $Z$ will crystallize out of the solution?

75.75\text{ g}

303.00\text{ g}

50.50\text{ g}

0.75\text{ g}

Show answer & explanation

Answer: $75.75\text{ g}$

Answer

The mass of salt Z that will crystallize out of solution is

75.75\text{ g}

.

Converting the solubility at

80^\circ\text{C}

(

5.0\text{ mol dm}^{-3}

) into mass concentration yields

505.0\text{ g dm}^{-3}

. Subtracting the solubility at

25^\circ\text{C}

(

202.0\text{ g dm}^{-3}

) gives

303.0\text{ g dm}^{-3}

precipitated. Multiplying by the volume ratio (

250\text{ cm}^3 / 1000\text{ cm}^3 = 0.25

) yields

75.75\text{ g}

.

Step-by-Step Solution

1

Convert the solubility at

80^\circ\text{C}

from

\text{mol dm}^{-3}

to

\text{g dm}^{-3}

.

\text{Solubility at } 80^\circ\text{C} = 5.0\text{ mol dm}^{-3} \times 101.0\text{ g mol}^{-1} = 505.0\text{ g dm}^{-3}

.

Solubility values given in molarity must be multiplied by molar mass to obtain concentration in mass per unit volume.

2

Calculate the mass of solute precipitated per

\text{dm}^3

upon cooling to

25^\circ\text{C}

.

\Delta\text{Solubility} = 505.0\text{ g dm}^{-3} - 202.0\text{ g dm}^{-3} = 303.0\text{ g dm}^{-3}

.

The difference between solubilities at the higher and lower temperatures gives the mass of solute that cannot remain dissolved in

1\text{ dm}^3

of water.

3

Scale the precipitated mass to the given volume of

250\text{ cm}^3

.

\text{Mass crystallized} = 303.0\text{ g dm}^{-3} \times \frac{250\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 303.0\text{ g} \times 0.25 = 75.75\text{ g}

.

The solution volume is

250\text{ cm}^3

(

0.25\text{ dm}^3

), so the precipitated mass is proportional to this fractional volume.

Key Concept

Crystallization calculations from solubility temperature changes

Question 56Question →

A saturated solution of potassium chlorate ( $\text{KClO}_3$ ) at $20^\circ\text{C}$ contains $7.35\text{ g}$ of solute dissolved in $100\text{ g}$ of distilled water. What is the solubility of $\text{KClO}_3$ at $20^\circ\text{C}$ in $\text{mol/dm}^3$ ? [Molar masses: $\text{K} = 39, \text{Cl} = 35.5, \text{O} = 16$ ; density of water $= 1.00\text{ g/cm}^3$ ]

Show answer & explanation

Answer: 0.6

Answer

The solubility of potassium chlorate at

20^\circ\text{C}

is

0.6\text{ mol/dm}^3

.

To convert mass of salt in a given solvent volume into solubility in

\text{mol/dm}^3

, first calculate the molar mass of

\text{KClO}_3

(

122.5\text{ g/mol}

). The amount of salt in moles is

7.35 / 122.5 = 0.06\text{ mol}

. Since

100\text{ g}

of water equals

0.1\text{ dm}^3

, the concentration of the saturated solution is

0.06\text{ mol} / 0.1\text{ dm}^3 = 0.6\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KClO}_3

122.5\text{ g/mol}

Sum the relative atomic masses:

39 (\text{K}) + 35.5 (\text{Cl}) + 3 \times 16 (\text{O}) = 122.5\text{ g/mol}

.

2

Calculate the number of moles of solute

0.06\text{ mol}

Divide the given mass by the molar mass:

\frac{7.35\text{ g}}{122.5\text{ g/mol}} = 0.06\text{ mol}

.

3

Convert the mass of solvent to volume in

\text{dm}^3

0.1\text{ dm}^3

Water density is

1.00\text{ g/cm}^3

, so

100\text{ g} = 100\text{ cm}^3 = 0.1\text{ dm}^3

.

4

Determine the molar solubility

0.6\text{ mol/dm}^3

Divide moles of solute by volume of solvent in

\text{dm}^3

:

\frac{0.06\text{ mol}}{0.1\text{ dm}^3} = 0.6\text{ mol/dm}^3

.

Key Concept

Solubility in mol/dm³

Question 57Question →

The table below shows the solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water at two different temperatures:

Temperature ( $^\circ\text{C}$ )	Solubility ( $\text{g / } 100\text{ g } \text{H}_2\text{O}$ )
60	110.0
20	32.0

What mass of $\text{KNO}_3$ will deposit when $420.0\text{ g}$ of a saturated solution of $\text{KNO}_3$ at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

78.0\text{ g}

156.0\text{ g}

220.0\text{ g}

327.6\text{ g}

Show answer & explanation

Answer: $156.0\text{ g}$

Answer

156.0\text{ g}

At

60^\circ\text{C}

, every

210.0\text{ g}

of saturated solution contains

100.0\text{ g}

of water and

110.0\text{ g}

of solute. Thus,

420.0\text{ g}

of solution contains

200.0\text{ g}

of water and

220.0\text{ g}

of dissolved

\text{KNO}_3

. Upon cooling to

20^\circ\text{C}

, the solubility drops to

32.0\text{ g}

per

100\text{ g}

of water, so

200.0\text{ g}

of water can retain only

64.0\text{ g}

of solute. The excess solute that crystallizes out is

220.0\text{ g} - 64.0\text{ g} = 156.0\text{ g}

.

Step-by-Step Solution

1

Determine the composition of a saturated solution at

60^\circ\text{C}

At

60^\circ\text{C}

,

100.0\text{ g}

of water dissolves

110.0\text{ g}

of

\text{KNO}_3

, yielding

100.0 + 110.0 = 210.0\text{ g}

of saturated solution.

Solubility is expressed per

100\text{ g}

of water, so total solution mass is the sum of solvent and solute masses.

2

Calculate the mass of water and solute in

420.0\text{ g}

of saturated solution at

60^\circ\text{C}

Mass of water =

420.0 \times \frac{100.0}{210.0} = 200.0\text{ g}

. Mass of

\text{KNO}_3

=

420.0 \times \frac{110.0}{210.0} = 220.0\text{ g}

.

Scaling the ratio of components to the given solution mass.

3

Calculate the mass of solute remaining dissolved at

20^\circ\text{C}

At

20^\circ\text{C}

,

100.0\text{ g}

of water holds

32.0\text{ g}

of

\text{KNO}_3

. Therefore,

200.0\text{ g}

of water holds

2 \times 32.0 = 64.0\text{ g}

of

\text{KNO}_3

.

Solubility at the lower temperature determines the maximum solute that stays in solution.

4

Calculate the mass of salt crystallized (deposited)

Mass deposited =

220.0\text{ g} - 64.0\text{ g} = 156.0\text{ g}

.

Subtracting the solute remaining in solution from the initial mass of dissolved solute.

Key Concept

Solubility curves and fractional crystallization calculations

Estimated Time:1m 30s

Question 58Question →

A saturated solution of potassium chloride, $\text{KCl}$ (molar mass = $74.5\text{ g mol}^{-1}$ ), has a solubility of $4.0\text{ mol dm}^{-3}$ at $80^\circ\text{C}$ and $1.5\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . If $600\text{ cm}^3$ of this saturated solution at $80^\circ\text{C}$ is cooled to $20^\circ\text{C}$ , what mass of $\text{KCl}$ will crystallize out of the solution?

111.75\text{ g}

67.05\text{ g}

178.80\text{ g}

1.50\text{ g}

Show answer & explanation

Answer: $111.75\text{ g}$

Answer

The mass of

\text{KCl}

that crystallizes out of the solution is

111.75\text{ g}

.

The correct answer is

111.75\text{ g}

. Cooling

0.60\text{ dm}^3

of saturated solution from

80^\circ\text{C}

to

20^\circ\text{C}

reduces the solubility by

2.5\text{ mol dm}^{-3}

, causing

1.50\text{ mol}

of

\text{KCl}

to precipitate. Multiplying

1.50\text{ mol}

by the molar mass (

74.5\text{ g mol}^{-1}

) yields

111.75\text{ g}

.

Step-by-Step Solution

1

Convert the volume of the solution from

\text{cm}^3

to

\text{dm}^3

.

V = \frac{600\text{ cm}^3}{1000} = 0.60\text{ dm}^3

Concentration is given in

\text{mol dm}^{-3}

, so volume must be in

\text{dm}^3

.

2

Calculate the difference in solubility between

80^\circ\text{C}

and

20^\circ\text{C}

.

\Delta C = 4.0\text{ mol dm}^{-3} - 1.5\text{ mol dm}^{-3} = 2.5\text{ mol dm}^{-3}

This represents the number of moles of solute precipitated per

\text{dm}^3

of solution upon cooling.

3

Calculate the number of moles crystallized in

0.60\text{ dm}^3

of solution.

n = 2.5\text{ mol dm}^{-3} \times 0.60\text{ dm}^3 = 1.50\text{ mol}

Scaling the molar amount precipitated to the specified solution volume.

4

Convert moles of crystallized salt to mass in grams.

\text{Mass} = 1.50\text{ mol} \times 74.5\text{ g mol}^{-1} = 111.75\text{ g}

Using the relation

\text{mass} = \text{moles} \times \text{molar mass}

.

Key Concept

Crystallization from Saturated Solutions on Cooling

Estimated Time:1m 30s

Question 59Question →

The solubility of a salt, $\text{NaCl}$ , at $298\text{ K}$ is $6.0\text{ mol dm}^{-3}$ . What mass of $\text{NaCl}$ must be dissolved in $250\text{ cm}^3$ of water to form a saturated solution at this temperature? $[\text{Molar mass of NaCl} = 58.5\text{ g mol}^{-1}]$

1.50\text{ g}

87.75\text{ g}

351.00\text{ g}

14.63\text{ g}

Show answer & explanation

Answer: $87.75\text{ g}$

Answer

The mass of

\text{NaCl}

required to prepare a saturated solution is

87.75\text{ g}

.

A saturated solution contains the maximum mass of solute dissolved at a specified temperature. At

298\text{ K}

,

1.0\text{ dm}^3

of saturated solution requires

6.0\text{ mol}

of

\text{NaCl}

. For

250\text{ cm}^3

(

0.25\text{ dm}^3

), the required amount is

1.5\text{ mol}

. Multiplying

1.5\text{ mol}

by

58.5\text{ g mol}^{-1}

yields

87.75\text{ g}

.

Step-by-Step Solution

1

Convert the volume of water from

\text{cm}^3

to

\text{dm}^3

.

\text{Volume} = \frac{250\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.25\text{ dm}^3

Solubility is expressed per

\text{dm}^3

, so the volume must be in

\text{dm}^3

.

2

Calculate the amount of

\text{NaCl}

in moles required to saturate

0.25\text{ dm}^3

of water.

\text{Moles} = 6.0\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 1.5\text{ mol}

The number of moles is obtained by multiplying molar concentration by volume in

\text{dm}^3

.

3

Convert moles of

\text{NaCl}

to mass using its molar mass.

\text{Mass} = 1.5\text{ mol} \times 58.5\text{ g mol}^{-1} = 87.75\text{ g}

Mass is calculated by multiplying the amount in moles by the molar mass.

Key Concept

Calculating solute mass required for saturation using solubility in

\text{mol dm}^{-3}

and molar mass.

Question 60Question →

At $50^\circ\text{C}$ , $21.2\text{ g}$ of anhydrous sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3$ , is dissolved in $200\text{ cm}^3$ of distilled water to form a saturated solution. Calculate the solubility of the salt at this temperature in $\text{mol/dm}^3$ . $[\text{Na} = 23, \text{C} = 12, \text{O} = 16]$

Fill in the blanks below

The solubility of

\text{Na}_2\text{CO}_3

at

50^\circ\text{C}

is

\text{mol/dm}^3

.

Show answer & explanation

Answer

The solubility of

\text{Na}_2\text{CO}_3

at

50^\circ\text{C}

is

1.00\text{ mol/dm}^3

.

To find the solubility in

\text{mol/dm}^3

, first calculate the molar mass of

\text{Na}_2\text{CO}_3

:

2(23) + 12 + 3(16) = 106\text{ g/mol}

. Next, convert

21.2\text{ g}

of salt into moles:

21.2 / 106 = 0.20\text{ mol}

. Then, convert

200\text{ cm}^3

of solvent into

\text{dm}^3

:

200 / 1000 = 0.20\text{ dm}^3

. Dividing moles by volume in

\text{dm}^3

gives

0.20 / 0.20 = 1.00\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{Na}_2\text{CO}_3

2(23) + 12 + 3(16) = 106\text{ g/mol}

Molar mass is required to convert mass of solute to moles.

2

Calculate the number of moles of

\text{Na}_2\text{CO}_3

dissolved

\frac{21.2\text{ g}}{106\text{ g/mol}} = 0.20\text{ mol}

Determines the mole quantity of solute present in solution.

3

Convert the volume of water from

\text{cm}^3

to

\text{dm}^3

\frac{200\text{ cm}^3}{1000} = 0.20\text{ dm}^3

Solubility in molarity requires volume in cubic decimeters.

4

Calculate solubility in

\text{mol/dm}^3

\frac{0.20\text{ mol}}{0.20\text{ dm}^3} = 1.00\text{ mol/dm}^3

Solubility is the number of moles of solute per

\text{dm}^3

of solvent.

Key Concept

Solubility Calculation in Moles per Cubic Decimeter

Air, Water and Solubility

Answer

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