Air, Water and Solubility Practice Questions — JAMB UTME — Page 4

Question 61Question →

The solubility of lead(II) trioxonitrate(V), $\text{Pb(NO}_3\text{)}_2$ , in water is $3.5\text{ mol dm}^{-3}$ at $80^\circ\text{C}$ and $1.5\text{ mol dm}^{-3}$ at $30^\circ\text{C}$ . What mass of $\text{Pb(NO}_3\text{)}_2$ will crystallize out when $250\text{ cm}^3$ of its saturated solution is cooled from $80^\circ\text{C}$ to $30^\circ\text{C}$ ? [Molar mass of $\text{Pb(NO}_3\text{)}_2 = 331\text{ g mol}^{-1}$ ]

165.5\text{ g}

0.50\text{ g}

662.0\text{ g}

289.6\text{ g}

Show answer & explanation

Answer: $165.5\text{ g}$

Answer

The mass of

\text{Pb(NO}_3\text{)}_2

that will crystallize out is

165.5\text{ g}

.

Subtracting the molar solubility at

30^\circ\text{C}

from that at

80^\circ\text{C}

gives a precipitation rate of

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

. Multiplying by the molar mass (

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

) gives

662.0\text{ g}

per

\text{dm}^3

. Scaling this value for

250\text{ cm}^3

(

0.25\text{ dm}^3

) yields

165.5\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in molar solubility between

80^\circ\text{C}

and

30^\circ\text{C}

.

\text{Molar solubility difference} = 3.5\text{ mol dm}^{-3} - 1.5\text{ mol dm}^{-3} = 2.0\text{ mol dm}^{-3}

This determines the amount of solute in moles that precipitates out per

\text{dm}^3

of solvent.

2

Calculate the mass of solute that crystallizes out from

1\text{ dm}^3

(

1000\text{ cm}^3

).

\text{Mass crystallized in } 1000\text{ cm}^3 = 2.0\text{ mol dm}^{-3} \times 331\text{ g mol}^{-1} = 662.0\text{ g dm}^{-3}

Converting the solubility difference from moles to grams using the given molar mass.

3

Scale the mass to the given solution volume of

250\text{ cm}^3

.

\text{Mass crystallized} = 662.0\text{ g} \times \frac{250\text{ cm}^3}{1000\text{ cm}^3} = 165.5\text{ g}

The volume provided (

250\text{ cm}^3

) is one-quarter of

1000\text{ cm}^3

(

1\text{ dm}^3

).

Key Concept

Mass of solute precipitated upon cooling a saturated solution

Question 62Question →

At $25^\circ\text{C}$ , $16.4\text{ g}$ of anhydrous calcium trioxonitrate(V), $\text{Ca(NO}_3\text{)}_2$ , is dissolved in $250\text{ cm}^3$ of distilled water to form a saturated solution. What is the solubility of the salt in $\text{mol/dm}^3$ at $25^\circ\text{C}$ ? $[\text{Ca} = 40, \text{N} = 14, \text{O} = 16]$

0.40\text{ mol/dm}^3

65.60\text{ mol/dm}^3

0.10\text{ mol/dm}^3

0.64\text{ mol/dm}^3

Show answer & explanation

Answer: $0.40\text{ mol/dm}^3$

Answer

The solubility of calcium trioxonitrate(V) at

25^\circ\text{C}

is

0.40\text{ mol/dm}^3

.

The correct answer is obtained by calculating the molar mass of calcium trioxonitrate(V) (

164\text{ g/mol}

), finding the moles dissolved (

0.10\text{ mol}

), and dividing by the solvent volume in cubic decimetres (

0.25\text{ dm}^3

), giving

0.40\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{Ca(NO}_3\text{)}_2

\text{Molar Mass} = 40 + 2(14 + 3 \times 16) = 40 + 2(62) = 164\text{ g/mol}

Molar mass is needed to convert the mass of the salt into moles.

2

Calculate the number of moles of

\text{Ca(NO}_3\text{)}_2

\text{Moles} = \frac{16.4\text{ g}}{164\text{ g/mol}} = 0.10\text{ mol}

Solubility in

\text{mol/dm}^3

requires the amount of solute in moles.

3

Convert the volume of distilled water from

\text{cm}^3

to

\text{dm}^3

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

Concentration units are expressed per cubic decimetre (

\text{dm}^3

).

4

Determine the molar solubility

\text{Solubility} = \frac{0.10\text{ mol}}{0.25\text{ dm}^3} = 0.40\text{ mol/dm}^3

Solubility in molar concentration is moles of solute divided by volume of solvent in

\text{dm}^3

.

Key Concept

Solubility Calculations in Molar Concentration

Question 63Question →

When white anhydrous copper(II) tetraoxosulfate(VI) powder is exposed to moist air, it turns blue by absorbing water vapor without forming a solution. Under the same conditions, solid sodium hydroxide pellets absorb moisture from the atmosphere until they completely dissolve to form a liquid solution. Which terms correctly classify the behavior of anhydrous copper(II) tetraoxosulfate(VI) and solid sodium hydroxide, respectively?

Hygroscopic and deliquescent

Deliquescent and hygroscopic

Efflorescent and deliquescent

Hygroscopic and efflorescent

Show answer & explanation

Answer: Hygroscopic and deliquescent

Answer

Anhydrous copper(II) tetraoxosulfate(VI) is hygroscopic, while solid sodium hydroxide is deliquescent.

Hygroscopic substances absorb moisture from the atmosphere without dissolving or forming a liquid solution, as seen with anhydrous copper(II) tetraoxosulfate(VI) turning blue (

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

). Deliquescent substances, such as solid sodium hydroxide (

\text{NaOH}

), absorb moisture from the atmosphere and dissolve in that absorbed water to form a saturated solution.

Step-by-Step Solution

1

Analyze the behavior of anhydrous copper(II) tetraoxosulfate(VI) in moist air.

It absorbs atmospheric water vapor to form hydrated copper(II) tetraoxosulfate(VI) (blue) without forming a liquid solution. Substances that absorb atmospheric moisture without dissolving are classified as hygroscopic.

Hygroscopy involves water absorption without phase change into a liquid solution.

2

Analyze the behavior of solid sodium hydroxide in moist air.

It absorbs sufficient atmospheric water to dissolve entirely and form a saturated solution. Substances that absorb water from the air and dissolve in it are classified as deliquescent.

Deliquescence occurs when a substance absorbs water until its vapor pressure matches atmospheric humidity, dissolving the solid.

3

Match the respective terms in order.

Hygroscopic and deliquescent.

The question asks for the behavior of anhydrous copper(II) tetraoxosulfate(VI) followed by sodium hydroxide.

Key Concept

Distinction between Hygroscopy, Deliquescence, and Efflorescence

Question 64Question →

A sample of hydrated sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ , is heated strongly in a crucible until a constant mass is reached. If the salt loses $62.94\%$ of its initial mass as water vapor during heating, what is the integer value of $x$ ?

[Relative atomic masses: $\text{Na} = 23$ , $\text{C} = 12$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 10

Answer

The integer value of

x

is 10.

Heating hydrated sodium trioxocarbonate(IV) drives off all water of crystallization as steam. Since

62.94\%

of the total mass is lost, water accounts for

62.94\%

of the molar mass of

\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}

. Solving

\frac{18x}{106 + 18x} = 0.6294

yields

x = 10

, representing decahydrate crystals,

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Step-by-Step Solution

1

Calculate the formula masses of the anhydrous salt

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

and water

\text{H}_2\text{O}

.

Molar mass of

\text{Na}_2\text{CO}_3 = 106\text{ g/mol}

and molar mass of

\text{H}_2\text{O} = 18\text{ g/mol}

.

Establishing the molar component masses is required to determine the percentage ratio of water of crystallization to the total mass of the hydrated salt.

2

Set up an algebraic ratio relating the mass of water lost to the total hydrated mass using the given percentage.

\frac{18x}{106 + 18x} = 0.6294

.

Heating to constant mass removes all water of crystallization, meaning the mass lost corresponds to the

x\text{H}_2\text{O}

component of the hydrated crystal.

3

Solve the algebraic equation for

x

.

x = 10

.

Isolating

x

yields the exact stoichiometric coefficient of water molecules per mole of hydrated salt.

Key Concept

Quantitative determination of water of crystallization using percentage mass loss.

Question 65Question →

A $600\text{ cm}^3$ sample of air contaminated with sulphur(IV) oxide ( $\text{SO}_2$ ) gas was passed through an excess aqueous solution of sodium hydroxide to absorb all the $\text{SO}_2$ , reducing the volume of the gas sample to $576\text{ cm}^3$ . The remaining gas mixture was then passed over excess heated copper turnings to remove oxygen gas, after which the unreacted gas volume measured $456\text{ cm}^3$ under the same conditions of temperature and pressure. What is the percentage by volume of sulphur(IV) oxide in the original air sample?

Show answer & explanation

Answer: 4

Answer

The percentage by volume of sulphur(IV) oxide in the original air sample is 4.0%.

Sodium hydroxide (

\text{NaOH}

) selectively absorbs acidic pollutant gases such as sulphur(IV) oxide (

\text{SO}_2

). The volume decrease from

600\text{ cm}^3

to

576\text{ cm}^3

indicates that

24\text{ cm}^3

of

\text{SO}_2

was absorbed. Dividing

24\text{ cm}^3

by the original sample volume of

600\text{ cm}^3

and multiplying by 100 gives

4.0\%

.

Step-by-Step Solution

1

Calculate the volume of sulphur(IV) oxide gas absorbed by the sodium hydroxide solution.

Volume of

\text{SO}_2 = 600\text{ cm}^3 - 576\text{ cm}^3 = 24\text{ cm}^3

.

Sodium hydroxide reacts with acidic oxide pollutants like

\text{SO}_2

, causing a reduction in gas volume equal to the volume of

\text{SO}_2

present.

2

Calculate the percentage composition by volume relative to the total initial sample.

\text{Percentage of } \text{SO}_2 = \left(\frac{24\text{ cm}^3}{600\text{ cm}^3}\right) \times 100\% = 4.0\%.

The volumetric percentage is determined by expressing the volume of the target gas component over the total volume of the original air sample.

Key Concept

Volumetric determination of air composition and gaseous pollutants.

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