All practice questions

Question 481Question →

A $500\text{ cm}^3$ sample of air collected near an industrial plant containing nitrogen ( $\text{N}_2$ ), oxygen ( $\text{O}_2$ ), carbon dioxide ( $\text{CO}_2$ ), and sulfur dioxide ( $\text{SO}_2$ ) pollutant was passed through concentrated potassium hydroxide ( $\text{KOH}$ ) solution, reducing the gas volume to $485\text{ cm}^3$ . The residual gas mixture was subsequently passed through excess alkaline solution of pyrogallol, resulting in a final volume of $383\text{ cm}^3$ . What is the percentage by volume of oxygen in the original air sample?

Show answer & explanation

Answer: 20.4

Answer

20.4%

Concentrated potassium hydroxide (KOH) absorbs the acidic gas pollutants (carbon dioxide and sulfur dioxide), causing an initial volume contraction of

15\text{ cm}^3

. Alkaline pyrogallol then absorbs elemental oxygen gas (

\text{O}_2

), causing a further contraction from

485\text{ cm}^3

to

383\text{ cm}^3

, which corresponds to

102\text{ cm}^3

of

\text{O}_2

. Dividing this volume of oxygen by the original total sample volume of

500\text{ cm}^3

and multiplying by

100\%

yields

20.4\%

.

Step-by-Step Solution

1

Identify the volume reduction caused by potassium hydroxide (KOH)

KOH absorbs acidic gases CO₂ and SO₂:

500\text{ cm}^3 - 485\text{ cm}^3 = 15\text{ cm}^3

.

Potassium hydroxide is an alkaline reagent that selectively absorbs acidic oxides present in polluted air.

2

Determine the volume of oxygen gas absorbed by alkaline pyrogallol

Alkaline pyrogallol absorbs O₂:

485\text{ cm}^3 - 383\text{ cm}^3 = 102\text{ cm}^3

.

Alkaline pyrogallol is a specific quantitative reagent used to absorb unreacted oxygen gas.

3

Calculate the percentage volume of oxygen in the initial sample

\left(\frac{102\text{ cm}^3}{500\text{ cm}^3}\right) \times 100\% = 20.4\%

.

The volume percentage of a component in air is the ratio of its volume to the total initial unreacted air sample volume multiplied by 100.

Key Concept

Quantitative volumetric determination of atmospheric components and gaseous pollutants using selective absorbents

Estimated Time:2m 0s

Question 482Question →

A uniform horizontal plank $AB$ of length $5.0\text{ m}$ and mass $40\text{ kg}$ rests on two smooth supports, $C$ and $D$ , located $1.0\text{ m}$ from end $A$ and $1.0\text{ m}$ from end $B$ respectively. What is the maximum mass, in kilograms, of an object that can be placed at end $A$ without causing the plank to tilt?

Show answer & explanation

Answer: 60

Answer

The maximum mass that can be placed at end A without tilting the plank is 60 kg.

Just before tilting, the plank rotates around support C, causing the normal reaction at support D to drop to zero. Equating the anticlockwise moment of the added mass at end A about support C (

(m \cdot g) \times 1.0\text{ m}

) to the clockwise moment of the plank's weight about support C (

(40 \cdot g) \times 1.5\text{ m}

) yields

m = 60\text{ kg}

.

Step-by-Step Solution

1

Identify the tipping condition and pivot point

Support C acts as the pivot; the reaction at support D becomes zero (

R_D = 0

).

When extra weight is added at end A, the plank rotates about C and lifts off support D.

2

Calculate perpendicular distances from the pivot C

Distance to added mass

m

=

1.0\text{ m}

; Distance to plank's center of mass =

2.5\text{ m} - 1.0\text{ m} = 1.5\text{ m}

.

The weight of a uniform beam acts at its midpoint (2.5 m from either end).

3

Equate clockwise and counterclockwise moments about C

m \times g \times 1.0\text{ m} = 40\text{ kg} \times g \times 1.5\text{ m}

, giving

m = 60\text{ kg}

.

For the plank to remain balanced just before tipping, total anticlockwise moment must equal total clockwise moment.

Key Concept

Rotational Equilibrium and Tilting of Rigid Bodies

Estimated Time:1m 30s

Question 483Question →

What volume of hydrogen gas, measured at STP, is produced when $4.8\text{ g}$ of magnesium ribbon reacts completely with excess dilute tetraoxosulfate(VI) acid according to the chemical equation below?

Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

$[Mg = 24, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

Show answer & explanation

Answer: 4.48

Answer

The volume of hydrogen gas produced at STP is

4.48\text{ dm}^3

.

From the stoichiometric relationship in the balanced reaction

Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

,

1\text{ mol}

of

Mg

(

24\text{ g}

) produces

1\text{ mol}

of

H_2

gas (

22.4\text{ dm}^3

at STP). For

4.8\text{ g}

of

Mg

, the number of moles is

\frac{4.8}{24} = 0.20\text{ mol}

. Multiplying by the molar gas volume gives

0.20 \times 22.4 = 4.48\text{ dm}^3

of

H_2

gas.

Step-by-Step Solution

1

Calculate the amount in moles of magnesium (

Mg

) reacted.

n(Mg) = \frac{4.8\text{ g}}{24\text{ g mol}^{-1}} = 0.20\text{ mol}

Dividing given mass by relative atomic mass yields the number of moles.

2

Determine the amount in moles of hydrogen gas (

H_2

) produced.

n(H_2) = 0.20\text{ mol}

The balanced chemical equation shows a

1:1

stoichiometric molar ratio between

Mg

and

H_2

.

3

Calculate the volume of

H_2

gas at STP.

V(H_2) = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

At standard temperature and pressure (STP), one mole of any gas occupies

22.4\text{ dm}^3

.

Key Concept

Mass-Volume stoichiometric calculation at STP

Question 484Question →

In a Young's double-slit experiment, the separation between two narrow slits is $0.40\text{ mm}$ and the interference pattern is observed on a screen placed $1.20\text{ m}$ away from the slits. If the distance between consecutive bright fringes on the screen is $1.80\text{ mm}$ , what is the wavelength of the light used in nanometers ( $\text{nm}$ )?

Show answer & explanation

Answer: 600

Answer

The wavelength of the light used is

600\text{ nm}

.

Using the Young's double-slit fringe spacing relation

\beta = \frac{\lambda D}{d}

, rearranging yields

\lambda = \frac{\beta d}{D}

. Substituting

\beta = 1.80 \times 10^{-3}\text{ m}

,

d = 4.0 \times 10^{-4}\text{ m}

, and

D = 1.20\text{ m}

gives

\lambda = 6.00 \times 10^{-7}\text{ m}

, which corresponds to

600\text{ nm}

.

Step-by-Step Solution

1

Convert given physical quantities into standard SI units (meters).

Slit separation

d = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}

, distance to screen

D = 1.20\text{ m}

, and fringe spacing

\beta = 1.80\text{ mm} = 1.80 \times 10^{-3}\text{ m}

.

Standard SI units ensure accuracy when applying wave speed and distance equations.

2

Write the Young's double-slit formula relating fringe width to wavelength.

\beta = \frac{\lambda D}{d}

This relationship defines the spatial period of interference fringes on a screen.

3

Rearrange the equation to isolate the wavelength

\lambda

.

\lambda = \frac{\beta d}{D}

The unknown parameter to solve for is the wavelength of the monochromatic source.

4

Substitute the numerical values and convert the final result to nanometers.

$\lambda = \frac{1.80 \times 10^{-3}\text{ m} \times 4.0 \times 10^{-4}\text{ m}}{1.20\text{ m}} = 6.00 \times 10^{-7}\text{ m} = 600\text{ nm}$

Multiply meters by

10^9

to express the wavelength in nanometers.

Key Concept

Young's Double-Slit Interference Fringe Spacing

Question 485Question →

In a metre bridge experiment, a standard resistor of $4.0\text{ }\Omega$ is connected in the left gap and an unknown resistor $R$ is connected in the right gap. When balanced, the balance point is found at a distance of $40.0\text{ cm}$ from the left end of the $100.0\text{ cm}$ bridge wire. What is the value of the unknown resistance $R$ in ohms?

Show answer & explanation

Answer: 6

Answer

The value of the unknown resistance

R

is

6.0\text{ }\Omega

.

The metre bridge operates on the Wheatstone bridge principle. At balance, the ratio of the resistance in the left gap to the length of the left segment equals the ratio of the resistance in the right gap to the length of the right segment. Substituting

4.0\text{ }\Omega

for the left gap and

40.0\text{ cm}

and

60.0\text{ cm}

for the two wire lengths yields

R = 6.0\text{ }\Omega

.

Step-by-Step Solution

1

Determine the length of the wire segment corresponding to the right gap.

l_2 = 100.0\text{ cm} - 40.0\text{ cm} = 60.0\text{ cm}

.

The total length of a standard metre bridge wire is

100.0\text{ cm}

.

2

Apply the Wheatstone bridge principle for the metre bridge balance condition.

\frac{R_{\text{left}}}{l_1} = \frac{R}{l_2} \implies R = R_{\text{left}} \times \frac{l_2}{l_1}

.

At balance, the potential drop per unit length across the two wire segments is proportional to their respective lengths.

3

Calculate the magnitude of the unknown resistor

R

.

R = 4.0 \times \frac{60.0}{40.0} = 6.0\text{ }\Omega

.

Multiplying and simplifying yields the exact resistance value.

Key Concept

Metre Bridge Principle (Wheatstone Bridge)

Question 486Question →

Monochromatic light of wavelength $600\text{ nm}$ is incident normally on a diffraction grating having $500\text{ lines per mm}$ . What is the angle of diffraction, in degrees, for the first-order principal maximum?

Show answer & explanation

Answer: 17.5

Answer

The angle of diffraction for the first-order principal maximum is

17.5^\circ

.

Using the grating equation

d \sin \theta = n \lambda

, the slit separation is

d = \frac{10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}

. For

n = 1

and

\lambda = 6.00 \times 10^{-7}\text{ m}

, we get

\sin \theta = \frac{6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.30

. Taking

\arcsin(0.30)

gives approximately

17.5^\circ

.

Step-by-Step Solution

1

Calculate the grating element (slit spacing)

d

d = 2.00 \times 10^{-6}\text{ m}

Grating spacing

d

is the reciprocal of the line density

N = 500\text{ lines/mm} = 500,000\text{ lines/m}

.

2

Apply the diffraction grating equation

d \sin \theta = n \lambda

\sin \theta = 0.30

For the first-order maximum (

n = 1

),

\sin \theta = \frac{1 \times 600 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.30

.

3

Find the angle

\theta

by taking the inverse sine

\theta = 17.5^\circ

\arcsin(0.30) \approx 17.46^\circ

, which rounds to

17.5^\circ

.

Key Concept

Diffraction Grating Equation for Principal Maxima

Estimated Time:1m 30s

Question 487Question →

A DC supply with an electromotive force (e.m.f.) of $24.0\text{ V}$ and an internal resistance of $2.0\text{ }\Omega$ is connected across three identical $12.0\text{ }\Omega$ resistors connected in parallel. What is the terminal potential difference across the battery in volts?

Show answer & explanation

Answer: 16

Answer

The terminal potential difference across the battery is

16.0\text{ V}

.

The three identical

12.0\text{ }\Omega

resistors in parallel combine to yield an equivalent external resistance of

4.0\text{ }\Omega

. Adding the battery's internal resistance of

2.0\text{ }\Omega

gives a total circuit resistance of

6.0\text{ }\Omega

. The total current drawn from the battery is

I = \frac{24.0\text{ V}}{6.0\text{ }\Omega} = 4.0\text{ A}

. The terminal potential difference is the voltage drop across the external circuit,

V = 4.0\text{ A} \times 4.0\text{ }\Omega = 16.0\text{ V}

.

Step-by-Step Solution

1

Find the equivalent external resistance of the three parallel resistors.

R_p = 4.0\text{ }\Omega

Three identical resistors

R = 12.0\text{ }\Omega

connected in parallel have an equivalent resistance of

R_p = \frac{12.0}{3} = 4.0\text{ }\Omega

.

2

Find the total circuit resistance by adding internal resistance to the parallel combination.

R_{total} = 6.0\text{ }\Omega

Internal resistance

r = 2.0\text{ }\Omega

acts in series with the external parallel combination:

R_{total} = R_p + r = 4.0 + 2.0 = 6.0\text{ }\Omega

.

3

Calculate the total current supplied by the cell.

I = 4.0\text{ A}

Using the circuit formula

I = \frac{E}{R_{total}}

, we divide the e.m.f. of

24.0\text{ V}

by the total resistance of

6.0\text{ }\Omega

.

4

Compute the terminal potential difference across the cell.

V = 16.0\text{ V}

The potential drop across the external parallel network is

V = I R_p = 4.0\text{ A} \times 4.0\text{ }\Omega = 16.0\text{ V}

, which equals

E - Ir = 24.0\text{ V} - (4.0\text{ A} \times 2.0\text{ }\Omega) = 16.0\text{ V}

.

Key Concept

Terminal Potential Difference and Internal Resistance

Estimated Time:1m 30s

Question 488Question →

The solubility of copper(II) tetraoxosulfate(VI), $\text{CuSO}_4$ , at $60^\circ\text{C}$ and $20^\circ\text{C}$ is $40.0\text{ g}$ and $21.0\text{ g}$ per $100\text{ g}$ of water respectively. Calculate the mass of $\text{CuSO}_4$ in grams that will crystallize out of solution when a saturated solution containing $250\text{ g}$ of water is cooled from $60^\circ\text{C}$ to $20^\circ\text{C}$ .

Show answer & explanation

Answer: 47.5

Answer

The mass of

\text{CuSO}_4

that crystallizes out of solution is

47.5\text{ g}

.

Subtracting the solubility at

20^\circ\text{C}

(

21.0\text{ g}

) from the solubility at

60^\circ\text{C}

(

40.0\text{ g}

) yields

19.0\text{ g}

of

\text{CuSO}_4

deposited per

100\text{ g}

of water. Multiplying by the ratio of actual solvent mass to reference solvent mass (

250\text{ g} / 100\text{ g} = 2.5

) gives

47.5\text{ g}

.

Step-by-Step Solution

1

Determine the mass of

\text{CuSO}_4

deposited per

100\text{ g}

of water on cooling.

40.0\text{ g} - 21.0\text{ g} = 19.0\text{ g}

per

100\text{ g}

of water.

The mass of solute precipitated per

100\text{ g}

of solvent is equal to the difference in solubility between the higher and lower temperatures.

2

Calculate the mass of solute deposited for

250\text{ g}

of water.

19.0\text{ g} \times \frac{250\text{ g}}{100\text{ g}} = 47.5\text{ g}

.

The amount of solute crystallized out is directly proportional to the total mass of solvent present.

Key Concept

Crystallization and Mass of Solute Deposited on Cooling

Question 489Question →

An electric iron rated at $1200\,\text{W}$ is operated for $5\,\text{hours}$ each day for a period of $30\,\text{days}$ . If electrical energy costs $\text{₦}20.00$ per kilowatt-hour ( $\text{kWh}$ ), what is the total cost of electricity consumed by the iron over this period in Naira ( $\text{₦}$ )?

Show answer & explanation

Answer: 3600

Answer

The total cost of electricity consumed by the iron over the 30-day period is 3600 Naira.

To find the cost of electrical energy in commercial units, express power in kilowatts (

1.2\,\text{kW}

) and time in total hours (

150\,\text{h}

). The energy consumed is

1.2 \times 150 = 180\,\text{kWh}

. At a tariff of

\text{₦}20.00

per

\text{kWh}

, the total cost is

180 \times 20 = 3600\,\text{Naira}

.

Step-by-Step Solution

1

Convert the power rating of the appliance from watts to kilowatts

P = \frac{1200\,\text{W}}{1000} = 1.2\,\text{kW}

Commercial energy consumption is calculated in kilowatt-hours (kWh), requiring power in kilowatts.

2

Calculate the total operating time in hours

t = 5\,\text{hours/day} \times 30\,\text{days} = 150\,\text{hours}

The usage duration across the month must be expressed in total hours.

3

Determine electrical energy consumed in kWh

E = P \times t = 1.2\,\text{kW} \times 150\,\text{h} = 180\,\text{kWh}

Energy is the product of power in kilowatts and time in hours.

4

Calculate total cost of energy consumed

\text{Total Cost} = 180\,\text{kWh} \times \text{₦}20.00/\text{kWh} = \text{₦}3600

Total cost is found by multiplying energy in kWh by the unit tariff rate.

Key Concept

Commercial electrical energy unit (kWh) and billing calculation

Question 490Question →

A rigid metallic cylinder contains a fixed mass of gas at an initial pressure of $1.2\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the gas is heated until its temperature reaches $177^\circ\text{C}$ while the volume remains constant, what is the final pressure of the gas in atmospheres?

Show answer & explanation

Answer: 1.8

Answer

The final pressure of the gas is

1.8\text{ atm}

.

Converting the temperatures to Kelvin (

T_1 = 300\text{ K}

and

T_2 = 450\text{ K}

) and applying Gay-Lussac's Pressure Law

\frac{P_1}{T_1} = \frac{P_2}{T_2}

gives

P_2 = 1.2 \times \frac{450}{300} = 1.8\text{ atm}

.

Step-by-Step Solution

1

Convert given temperatures to the Kelvin scale

T_1 = 300\text{ K}

and

T_2 = 450\text{ K}

Gas laws require absolute temperature values measured in Kelvin (

T_K = t_C + 273

).

2

Set up Gay-Lussac's Pressure Law proportion

\frac{P_1}{T_1} = \frac{P_2}{T_2}

P_2 = P_1 \times \frac{T_2}{T_1} = 1.2 \times \frac{450}{300}

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

3

Perform the multiplication to determine the final pressure

P_2 = 1.8\text{ atm}

Multiplying

1.2

by the ratio

1.5

yields

1.8\text{ atm}

.

Key Concept

Pressure Law (Gay-Lussac's Law) states that for a given mass of gas at constant volume, the pressure is directly proportional to its absolute temperature in Kelvin (

P \propto T

).

Question 491Question →

Chief Kalu insured his commercial supermarket building, valued at $\text{N}50,000,000$ , against fire for a sum of $\text{N}35,000,000$ . The policy contains an average clause. If a fire outbreak damages the building causing a loss of $\text{N}10,000,000$ , what amount in Naira will the insurance company pay as indemnity?

Show answer & explanation

Answer: 7000000

Answer

7,000,000 Naira

Under the average clause in fire insurance policies, if a property is under-insured (insured for less than its true value), the insured is deemed to be their own insurer for the uninsured portion. Here, the building was insured for 70% of its full value (N35,000,000 / N50,000,000 = 0.70). Therefore, the insurance company pays only 70% of the actual loss incurred (70% of N10,000,000 = 7,000,000 Naira).

Step-by-Step Solution

1

Identify the property values given in the problem statement.

Actual Value = N50,000,000; Sum Insured = N35,000,000; Actual Loss = N10,000,000.

These figures are required to calculate indemnity under under-insurance.

2

Apply the average clause compensation formula.

Compensation = (Sum Insured / Actual Value) * Actual Loss

When property is insured for less than its full value and has an average clause, the insurer pays only a proportionate share of any loss.

3

Calculate the compensation payable.

Compensation = (35,000,000 / 50,000,000) * 10,000,000 = 7,000,000 Naira.

Since the owner insured 70% of the building's value, the insurer pays 70% of the actual loss suffered.

Key Concept

Average Clause in Insurance

Question 492Question →

At a magnetic observation station, the horizontal component of the Earth's magnetic field is $40\text{ }\mu\text{T}$ and the vertical component is $30\text{ }\mu\text{T}$ . What is the total magnetic field intensity of the Earth at this station in $\mu\text{T}$ ?

Show answer & explanation

Answer: 50

Answer

The total magnetic field intensity of the Earth at this station is

50\text{ }\mu\text{T}

.

The horizontal component (

B_h

) and vertical component (

B_v

) of the Earth's magnetic field act at right angles to each other. Therefore, the resultant total magnetic field intensity (

B

) is calculated using vector addition:

B = \sqrt{B_h^2 + B_v^2} = \sqrt{40^2 + 30^2} = 50\text{ }\mu\text{T}

.

Step-by-Step Solution

1

Identify the vector relationship between the horizontal and vertical components of the Earth's magnetic field.

B = \sqrt{B_h^2 + B_v^2}

, where

B_h = 40\text{ }\mu\text{T}

and

B_v = 30\text{ }\mu\text{T}

.

The horizontal and vertical components of the Earth's magnetic field are mutually perpendicular vector components.

2

Substitute the values into the formula and solve for total magnetic field intensity

B

.

B = \sqrt{(40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ }\mu\text{T}

.

Applying the Pythagorean theorem yields the magnitude of the resultant magnetic field vector.

Key Concept

Resolution of Earth's magnetic field into horizontal (

B_h

) and vertical (

B_v

) components.

Estimated Time:1m 0s

Question 493Question →

A rigid container holds a gas mixture containing $4.0\text{ g}$ of methane ( $\text{CH}_4$ ) and $14.0\text{ g}$ of nitrogen ( $\text{N}_2$ ). If the partial pressure exerted by methane in the mixture is $125\text{ kPa}$ , what is the total pressure of the gas mixture in $\text{kPa}$ ? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{N} = 14$ ]

Show answer & explanation

Answer: 375

Answer

The total pressure of the gas mixture is

375\text{ kPa}

.

The total pressure of

375\text{ kPa}

is determined by calculating the moles of methane (

0.25\text{ mol}

) and nitrogen (

0.50\text{ mol}

), giving a total of

0.75\text{ mol}

. Methane constitutes one-third (

1/3

) of the total moles, so its partial pressure is one-third of the total pressure. Dividing the partial pressure of methane (

125\text{ kPa}

) by its mole fraction (

1/3

) yields a total pressure of

375\text{ kPa}

.

Step-by-Step Solution

1

Calculate the amount of moles of each gas present in the mixture.

n_{\text{CH}_4} = \frac{4.0\text{ g}}{16\text{ g/mol}} = 0.25\text{ mol}

and

n_{\text{N}_2} = \frac{14.0\text{ g}}{28\text{ g/mol}} = 0.50\text{ mol}

.

Molar masses are determined from relative atomic masses:

\text{CH}_4 = 12 + 4(1) = 16\text{ g/mol}

and

\text{N}_2 = 2(14) = 28\text{ g/mol}

.

2

Calculate total moles and the mole fraction of methane.

n_{\text{total}} = 0.25 + 0.50 = 0.75\text{ mol}

;

X_{\text{CH}_4} = \frac{0.25\text{ mol}}{0.75\text{ mol}} = \frac{1}{3}

.

Mole fraction is the ratio of the number of moles of a specific gas component to the total number of moles in the gas mixture.

3

Apply Dalton's Law of Partial Pressures to find total pressure.

P_{\text{total}} = \frac{P_{\text{CH}_4}}{X_{\text{CH}_4}} = \frac{125\text{ kPa}}{1/3} = 375\text{ kPa}

.

According to Dalton's Law, the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure (

P_i = X_i \times P_{\text{total}}

).

Key Concept

Dalton's Law of Partial Pressures and Mole Fraction

Estimated Time:1m 30s

Question 494Question →

The solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water at $40^\circ\text{C}$ is $6.0\text{ mol dm}^{-3}$ . What mass of $\text{KNO}_3$ , in grams, is required to prepare a saturated solution in $250\text{ cm}^3$ of water at this temperature? (Molar mass of $\text{KNO}_3 = 101\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 151.5

Answer

The mass of

\text{KNO}_3

required to prepare a saturated solution in

250\text{ cm}^3

of water at

40^\circ\text{C}

is

151.5\text{ g}

.

To find the mass of solute required for saturation, convert the given volume of water to cubic decimeters (

250\text{ cm}^3 = 0.25\text{ dm}^3

). Multiply the volume by the molar solubility (

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

) to obtain the number of moles, then multiply by the molar mass (

1.5\text{ mol} \times 101\text{ g mol}^{-1} = 151.5\text{ g}

).

Step-by-Step Solution

1

Convert volume from

\text{cm}^3

to

\text{dm}^3

0.25\text{ dm}^3

Molar solubility is expressed per

\text{dm}^3

, so the volume of solvent must be in

\text{dm}^3

.

2

Determine moles of

\text{KNO}_3

needed for saturation

1.5\text{ mol}

Multiply molar solubility by the volume in

\text{dm}^3

.

3

Convert moles to mass in grams

151.5\text{ g}

Multiply moles by the molar mass of

\text{KNO}_3

(

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

).

Key Concept

Calculating solute mass for saturation using molar solubility and volume

Question 495Question →

A naturally occurring sample of neon gas consists of $90\%$ $^{20}\text{Ne}$ and $10\%$ $^{22}\text{Ne}$ . What is the relative atomic mass of neon in this sample?

Show answer & explanation

Answer: 20.2

Answer

The relative atomic mass of neon in the sample is 20.2.

The relative atomic mass of an element is calculated by summing the products of the mass number of each isotope and its fractional abundance:

\text{RAM} = (20 \times 0.90) + (22 \times 0.10) = 18.0 + 2.2 = 20.2

.

Step-by-Step Solution

1

Calculate the weighted contribution of

^{20}\text{Ne}

20 \times 0.90 = 18.0

The isotope

^{20}\text{Ne}

accounts for

90\%

of the sample.

2

Calculate the weighted contribution of

^{22}\text{Ne}

22 \times 0.10 = 2.2

The isotope

^{22}\text{Ne}

accounts for

10\%

of the sample.

3

Sum the weighted contributions to find the relative atomic mass

18.0 + 2.2 = 20.2

The relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes.

Key Concept

Calculation of Relative Atomic Mass from Isotopic Abundance

Question 496Question →

A solution of a monoprotic acid $HX$ contains $3.65\text{ g dm}^{-3}$ of the acid. If $25.0\text{ cm}^3$ of this acid solution neutralizes $20.0\text{ cm}^3$ of a $0.125\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution, what is the molar mass of the acid $HX$ in $\text{g mol}^{-3}$ ?

Show answer & explanation

Answer: 36.5

Answer

The molar mass of the acid

HX

is

36.5\text{ g mol}^{-3}

.

The reaction of monoprotic acid

HX

with

\text{NaOH}

follows a 1:1 stoichiometric ratio. Applying the titration formula

\frac{C_a V_a}{C_b V_b} = 1

yields an acid concentration of

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

. Dividing the mass concentration of

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

by this molarity gives a molar mass of

36.5\text{ g mol}^{-3}

.

Step-by-Step Solution

1

Determine the stoichiometric mole ratio between acid and base.

The mole ratio

n_a : n_b

for

HX

reacting with

\text{NaOH}

is

1 : 1

.

A monoprotic acid donates one proton per molecule to react with one mole of sodium hydroxide.

2

Calculate the molar concentration of the acid (

C_a

).

C_a = \frac{0.125\text{ mol dm}^{-3} \times 20.0\text{ cm}^3}{25.0\text{ cm}^3} = 0.100\text{ mol dm}^{-3}

.

Using the titration relation

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

allows determination of the acid molarity.

3

Calculate the molar mass of the acid.

\text{Molar mass} = \frac{\text{Mass concentration}}{\text{Molarity}} = \frac{3.65\text{ g dm}^{-3}}{0.100\text{ mol dm}^{-3}} = 36.5\text{ g mol}^{-3}

.

Molar mass is defined as mass of substance per mole.

Key Concept

Determination of molar mass using volumetric titration stoichiometry

Estimated Time:1m 30s

Question 497Question →

Consider the redox reaction between dichromate ions (

\text{Cr}_2\text{O}_7^{2-}

) and iron(II) ions (

\text{Fe}^{2+}

) in an acidic medium:

\text{Cr}_2\text{O}_7^{2-} + x\text{Fe}^{2+} + y\text{H}^+ \rightarrow 2\text{Cr}^{3+} + x\text{Fe}^{3+} + z\text{H}_2\text{O}

What is the stoichiometric coefficient

x

of

\text{Fe}^{2+}

when the ionic equation is completely balanced?

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient x of Fe²⁺ in the balanced redox reaction is 6.

In the reduction half-reaction, dichromate (

\text{Cr}_2\text{O}_7^{2-}

) contains two Cr atoms in the +6 oxidation state converting to two

\text{Cr}^{3+}

ions in the +3 state, which consumes 6 electrons. In the oxidation half-reaction, each

\text{Fe}^{2+}

ion loses 1 electron to form

\text{Fe}^{3+}

. To balance charge transfer, 6

\text{Fe}^{2+}

ions are needed for every 1

\text{Cr}_2\text{O}_7^{2-}

ion, making the stoichiometric coefficient

x

equal to 6.

Step-by-Step Solution

1

Determine the oxidation state changes for Chromium and Iron.

Chromium changes from +6 in

\text{Cr}_2\text{O}_7^{2-}

to +3 in

\text{Cr}^{3+}

, requiring 3 electrons per Chromium atom (

6e^-

total for two Cr atoms). Iron changes from +2 in

\text{Fe}^{2+}

to +3 in

\text{Fe}^{3+}

, releasing

1e^-

per Iron atom.

Identifying the number of electrons transferred in each half-reaction is required to balance the overall redox equation.

2

Balance the electron gain and loss.

The oxidation half-reaction (

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

) must be multiplied by 6 to balance the 6 electrons required by the dichromate ion.

The total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent.

3

Combine the half-reactions and read the coefficient

x

.

The balanced chemical equation is

\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}

, giving

x = 6

.

The coefficient

x

corresponds directly to the stoichiometric multiplier applied to

\text{Fe}^{2+}

.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Question 498Question →

A galvanic cell is constructed under standard conditions using the following two reduction half-reactions:

\text{Fe}^{3+}_{\text{(aq)}} + \text{e}^- \rightarrow \text{Fe}^{2+}_{\text{(aq)}}, \quad E^\circ = +0.77\text{ V}

\text{Sn}^{4+}_{\text{(aq)}} + 2\text{e}^- \rightarrow \text{Sn}^{2+}_{\text{(aq)}}, \quad E^\circ = +0.15\text{ V}

What is the standard cell potential ( $E^\circ_{\text{cell}}$ ) in volts for the spontaneous overall reaction?

Show answer & explanation

Answer: 0.62

Answer

The standard cell potential for the spontaneous reaction is +0.62 V.

In a spontaneous galvanic cell, reduction takes place at the cathode, which is the electrode with the higher standard reduction potential (+0.77 V for

\text{Fe}^{3+}/\text{Fe}^{2+}

). Oxidation occurs at the anode (+0.15 V for

\text{Sn}^{4+}/\text{Sn}^{2+}

). Substituting these values into

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

yields

+0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V}

.

Step-by-Step Solution

1

Identify the cathode and anode based on standard reduction potentials.

Cathode: Fe3+/Fe2+ half-cell (E° = +0.77 V); Anode: Sn4+/Sn2+ half-cell (E° = +0.15 V).

In a spontaneous galvanic cell, reduction occurs at the electrode with the more positive standard reduction potential.

2

Apply the standard electromotive force equation.

E°cell = E°cathode - E°anode = +0.77 V - (+0.15 V) = +0.62 V

The cell potential measures the overall potential difference between the reduction half-cell and oxidation half-cell.

Key Concept

Standard Cell Potential Calculation

Estimated Time:1m 30s

Question 499Question →

A $1.5\text{ mol}$ sample of carbon dioxide gas ( $\text{CO}_2$ ) is held in a container of volume $0.60\text{ dm}^3$ at a temperature of $300\text{ K}$ under a pressure of $50\text{ atm}$ . What is the compressibility factor ( $Z$ ) for the gas under these conditions? (Take $R = 0.082\text{ atm dm}^3\text{ mol}^{-1}\text{ K}^{-1}$ and express your answer to two decimal places.)

Show answer & explanation

Answer: 0.81

Answer

The compressibility factor (Z) for carbon dioxide under these conditions is 0.81.

The compressibility factor

Z

is defined as

Z = \frac{PV}{nRT}

. Substituting the given values yields

PV = 50 \times 0.60 = 30.0

and

nRT = 1.5 \times 0.082 \times 300 = 36.9

. Dividing

30.0

by

36.9

yields approximately

0.81

, which indicates negative deviation from ideality due to intermolecular attractive forces.

Step-by-Step Solution

1

Identify the given variables and the formula for the compressibility factor.

P = 50\text{ atm}

,

V = 0.60\text{ dm}^3

,

n = 1.5\text{ mol}

,

T = 300\text{ K}

,

R = 0.082\text{ atm dm}^3\text{ mol}^{-1}\text{ K}^{-1}

. Formula:

Z = \frac{PV}{nRT}

.

The compressibility factor

Z

measures the deviation of a real gas from ideal gas behavior (

Z = 1

for an ideal gas).

2

Calculate the actual pressure-volume product (

PV

).

PV = 50\text{ atm} \times 0.60\text{ dm}^3 = 30.0\text{ atm dm}^3

.

This represents the numerator in the compressibility ratio.

3

Calculate the theoretical ideal pressure-volume product (

nRT

).

nRT = 1.5\text{ mol} \times 0.082\text{ atm dm}^3\text{ mol}^{-1}\text{ K}^{-1} \times 300\text{ K} = 36.9\text{ atm dm}^3

.

This represents the expected

PV

value if the gas behaved ideally.

4

Compute the value of

Z

and round to two decimal places.

Z = \frac{30.0}{36.9} \approx 0.8130 \rightarrow 0.81

.

Dividing the observed

PV

by

nRT

gives

Z < 1

, indicating that intermolecular attractive forces dominate under these conditions.

Key Concept

Compressibility factor Z of real gases

Question 500Question →

Meridian Logistics Plc was registered with an authorized share capital of $5,000,000$ ordinary shares of $\text{₦}1.50$ each. The directors issued $3,000,000$ shares to the public and called up $\text{₦}1.00$ per share. If all called-up funds were received except a call of $\text{₦}0.20$ per share on $150,000$ shares, calculate the total paid-up share capital of the company in Naira.

Show answer & explanation

Answer: 2970000

Answer

The paid-up capital of Meridian Logistics Plc is ₦2,970,000.

Paid-up capital represents the actual cash received from shareholders for called-up shares. The called-up capital is

3,000,000 \text{ shares} \times \text{₦}1.00 = \text{₦}3,000,000

. Calls in arrears are

150,000 \text{ shares} \times \text{₦}0.20 = \text{₦}30,000

. Deducting calls in arrears from called-up capital gives

\text{₦}3,000,000 - \text{₦}30,000 = \text{₦}2,970,000

.

Step-by-Step Solution

1

Calculate Total Called-up Capital

₦3,000,000

Multiply issued shares by called-up value per share (3,000,000 shares × ₦1.00).

2

Calculate Calls in Arrears

₦30,000

Multiply defaulting shares by unpaid call per share (150,000 shares × ₦0.20).

3

Compute Paid-up Capital

₦2,970,000

Subtract Calls in Arrears from Total Called-up Capital (₦3,000,000 - ₦30,000).

Key Concept

Classification of Share Capital - Paid-up Capital Calculation

Estimated Time:1m 30s

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