All practice questions

Question 461Question →

BlueWave Marine Plc has an authorized share capital of $2,000,000$ ordinary shares of $₦5$ each. The company issued $80\%$ of its authorized shares to the public and subsequently called up $₦4$ per share. All shareholders paid the call in full except for holders of $50,000$ shares who defaulted on the payment. What is the total paid-up capital of the company in Naira?

Show answer & explanation

Answer: 6200000

Answer

The total paid-up capital of the company is

₦6,200,000

.

Paid-up capital is the portion of called-up capital that shareholders have actually paid into the business. First,

80\%

of the

2,000,000

authorized shares equals

1,600,000

issued shares. Calling up

₦4

per share yields a called-up amount of

₦6,400,000

. Since holders of

50,000

shares defaulted on

₦4

per share, the unpaid calls in arrears total

₦200,000

. Subtracting

₦200,000

from

₦6,400,000

results in a total paid-up capital of

₦6,200,000

.

Step-by-Step Solution

1

Determine the number of issued shares

1,600,000

shares

The company issued

80\%

of its

2,000,000

authorized shares (

0.80 \times 2,000,000 = 1,600,000

).

2

Calculate total called-up capital

₦6,400,000

The directors requested

₦4

per share across all

1,600,000

issued shares (

1,600,000 \times ₦4 = ₦6,400,000

).

3

Calculate calls in arrears

₦200,000

Holders of

50,000

shares failed to pay the requested

₦4

per share (

50,000 \times ₦4 = ₦200,000

).

4

Compute net paid-up capital

₦6,200,000

Paid-up capital represents actual cash received, which equals Called-Up Capital minus Calls in Arrears (

₦6,400,000 - ₦200,000 = ₦6,200,000

).

Key Concept

Paid-Up Capital and Calls in Arrears

Question 462Question →

A radioactive sample of an isotope has a half-life of $20\text{ minutes}$ . If the initial mass of the sample is $80\text{ g}$ , what mass of the isotope, in grams, will remain undecayed after $60\text{ minutes}$ ?

Show answer & explanation

Answer: 10

Answer

10 g of the isotope remains undecayed.

The correct answer is derived from the half-life formula

N_t = N_0 \times (1/2)^n

. Since

60\text{ minutes}

contains three

20\text{ minute}

half-life periods (

n = 3

), the remaining mass is

80\text{ g} \times (1/2)^3 = 80\text{ g} / 8 = 10\text{ g}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives

3 half-lives

Divide the total time elapsed (

60\text{ minutes}

) by the duration of one half-life (

20\text{ minutes}

).

2

Calculate the remaining mass after 3 half-lives

10 g

Apply the radioactive decay relation

N_t = N_0 \times \left(\frac{1}{2}\right)^n

, yielding

80 \times \left(\frac{1}{2}\right)^3 = 10\text{ g}

.

Key Concept

Radioactive Half-Life and Exponential Decay

Question 463Question →

A parallel plate capacitor with a capacitance of $12\text{ }\mu\text{F}$ is fully charged by connecting it across a $50\text{ V}$ direct current power source. What is the total electrostatic energy stored in the capacitor in millijoules ( $\text{mJ}$ )?

Show answer & explanation

Answer: 15

Answer

The total electrostatic energy stored in the capacitor is

15\text{ mJ}

.

The energy stored in a capacitor is given by

E = \frac{1}{2}CV^2

. Substituting

C = 12 \times 10^{-6}\text{ F}

and

V = 50\text{ V}

yields

E = \frac{1}{2} \times 12 \times 10^{-6} \times 2500 = 0.015\text{ J} = 15\text{ mJ}

.

Step-by-Step Solution

1

Convert given values to standard SI units.

C = 12 \times 10^{-6}\text{ F}

,

V = 50\text{ V}

Calculating in SI units ensures the resultant energy is in Joules.

2

Apply the formula for energy stored in a charged capacitor.

E = \frac{1}{2} C V^2

Work done during charging is stored as electrostatic potential energy in the electric field between the plates.

3

Substitute the values and convert Joules to millijoules.

E = \frac{1}{2} \times (12 \times 10^{-6}) \times 2500 = 0.015\text{ J} = 15\text{ mJ}

Multiplying Joules by

10^3

converts the result into millijoules.

Key Concept

Energy stored in a capacitor (

E = \frac{1}{2}CV^2

)

Question 464Question →

A $400\text{ cm}^3$ sample of polluted air containing nitrogen dioxide ( $\text{NO}_2$ ), carbon dioxide ( $\text{CO}_2$ ), and unpolluted air components is passed sequentially through two absorption reagents. First, passing the sample through concentrated sodium hydroxide ( $\text{NaOH}$ ) solution absorbs both acidic pollutants ( $\text{NO}_2$ and $\text{CO}_2$ ), reducing the volume of the gas by $40\text{ cm}^3$ . The remaining gas mixture is then passed through alkaline pyrogallol, where oxygen ( $\text{O}_2$ ) is completely absorbed, causing a further volume reduction of $75.6\text{ cm}^3$ . Assuming oxygen constitutes $21\%$ by volume of the unpolluted portion of the air sample, what is the percentage by volume of nitrogen dioxide ( $\text{NO}_2$ ) in the original polluted air sample?

Show answer & explanation

Answer: 8.5

Answer

The percentage by volume of nitrogen dioxide (NO₂) in the original polluted air sample is 8.5%.

The absorption of oxygen (75.6 cm³) by alkaline pyrogallol represents 21% of the unpolluted air fraction, giving an unpolluted volume of 360 cm³. The difference between the original sample volume (400 cm³) and unpolluted air volume (360 cm³) represents the total pollutant contraction (40 cm³). Subtracting the 6 cm³ CO₂ contribution leaves 34 cm³ of NO₂, which equals (34 / 400) * 100% = 8.5% by volume.

Step-by-Step Solution

1

Calculate the volume of the unpolluted portion of the air sample

360 cm³

Alkaline pyrogallol absorbs oxygen gas. Since oxygen constitutes 21% by volume of unpolluted air and 75.6 cm³ of O₂ was absorbed, the volume of unpolluted air is equal to 75.6 cm³ divided by 0.21.

2

Determine the total volume of polluted acidic gases (NO₂ and CO₂)

40 cm³

The total volume contraction when passed through concentrated NaOH is 40 cm³, as NaOH reacts with and absorbs acidic oxides such as NO₂ and CO₂.

3

Calculate the volume of nitrogen dioxide (NO₂) in the sample

34 cm³

Subtracting the unpolluted air volume (360 cm³) from the total sample volume (400 cm³) confirms that total acidic pollutant gas volume is 40 cm³. Deducting the background CO₂ component (6 cm³) leaves 34 cm³ of NO₂.

4

Calculate the percentage by volume of NO₂ in the original sample

8.5%

Dividing the volume of NO₂ (34 cm³) by the total sample volume (400 cm³) and multiplying by 100% yields 8.5%.

Key Concept

Quantitative determination of air composition and gaseous pollutants via selective volumetric absorption

Question 465Question →

A mixture of $25\text{ cm}^3$ of methane ( $CH_4$ ) and $60\text{ cm}^3$ of oxygen ( $O_2$ ) is sparked at constant temperature and pressure. Assuming water vapor condenses to liquid upon cooling to room temperature, what is the total volume of residual gas remaining in $\text{cm}^3$ ?

Show answer & explanation

Answer: 35

Answer

The total volume of residual gas remaining after cooling to room temperature is

35\text{ cm}^3

.

According to Gay-Lussac's law of combining volumes, gases react in simple numerical ratios equal to their stoichiometric coefficients at constant temperature and pressure. For the equation

CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

,

25\text{ cm}^3

of methane reacts completely with

50\text{ cm}^3

of oxygen to produce

25\text{ cm}^3

of

CO_2

gas. Since

60\text{ cm}^3

of oxygen was initially present,

10\text{ cm}^3

of oxygen remains unreacted. Liquid water occupies negligible volume compared to gases. Therefore, the total residual gaseous volume is the sum of unreacted oxygen and produced carbon(IV) oxide:

10\text{ cm}^3 + 25\text{ cm}^3 = 35\text{ cm}^3

.

Step-by-Step Solution

1

Write the balanced chemical equation for the combustion of methane gas.

CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

Gay-Lussac's law of combining volumes applies directly to the stoichiometric mole ratios of gaseous reactants and products.

2

Determine the volume of oxygen consumed and identify the excess reactant.

25\text{ cm}^3

of

CH_4

consumes

50\text{ cm}^3

of

O_2

, leaving

10\text{ cm}^3

of unreacted

O_2

.

The reaction stoichiometry requires 2 volumes of oxygen per 1 volume of methane.

3

Calculate the volume of gaseous carbon(IV) oxide produced.

Volume of

CO_2(g) = 25\text{ cm}^3

.

1 volume of methane produces 1 volume of carbon(IV) oxide gas.

4

Sum the volumes of all remaining gaseous species.

Total residual volume =

10\text{ cm}^3\text{ (excess } O_2) + 25\text{ cm}^3\text{ (formed } CO_2) = 35\text{ cm}^3

.

Water formed is liquid at room temperature and contributes negligibly to gaseous volume.

Key Concept

Gay-Lussac's Law of Combining Volumes and Stoichiometry of Gas Reactions

Question 466Question →

A saturated solution of potassium chloride ( $\text{KCl}$ ) contains $14.9\text{ g}$ of the salt dissolved in $100\text{ g}$ of water at $25^\circ\text{C}$ . What is the solubility of potassium chloride in $\text{mol/dm}^3$ at this temperature?

(Take density of water = $1.0\text{ g/cm}^3$ , relative atomic masses: $\text{K} = 39$ , $\text{Cl} = 35.5$ )

Show answer & explanation

Answer: 2

Answer

The solubility of potassium chloride at

25^\circ\text{C}

is

2.0\text{ mol/dm}^3

.

The correct calculated value is

2.0\text{ mol/dm}^3

. Molar mass of

\text{KCl} = 39 + 35.5 = 74.5\text{ g/mol}

. Number of moles of

\text{KCl} = \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}

. Volume of water solvent

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

. Therefore, solubility

= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KCl}

and convert mass of solute to moles.

Molar mass

= 74.5\text{ g/mol}

; Moles

= \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}

.

Solubility in

\text{mol/dm}^3

requires the quantity of solute to be expressed in moles rather than grams.

2

Convert the mass/volume of solvent into decimeters cubed (

\text{dm}^3

).

Volume

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

.

Molar solubility concentration is defined per

1.0\text{ dm}^3

of solution/solvent.

3

Compute the concentration in

\text{mol/dm}^3

by dividing moles of solute by volume of solvent in

\text{dm}^3

.

Solubility

= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3

.

Solubility in molarity equal to total moles divided by total volume in

\text{dm}^3

.

Key Concept

Calculating molar solubility in

\text{mol/dm}^3

from solute mass and solvent volume.

Question 467Question →

A sample of oxygen gas is collected over water at $27^\circ\text{C}$ and a total pressure of $755\text{ mmHg}$ . If the saturated vapor pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ , what is the partial pressure of the dry oxygen gas in $\text{mmHg}$ ?

Show answer & explanation

Answer: 730

Answer

The partial pressure of the dry oxygen gas is

730\text{ mmHg}

.

According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. When a gas is collected over water, the gas absorbs water vapor, so

P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapor}}

. To find the pressure of the dry oxygen, the water vapor pressure (

25\text{ mmHg}

) must be subtracted from the total barometric pressure (

755\text{ mmHg}

), giving

730\text{ mmHg}

.

Step-by-Step Solution

1

Identify Dalton's Law equation for a gas collected over water

P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water}}

When a gas is collected over water, it becomes saturated with water vapor. The total observed pressure is the sum of the partial pressure of the dry gas and the vapor pressure of water (aqueous tension).

2

Subtract the aqueous tension from the total pressure

P_{\text{dry gas}} = 755\text{ mmHg} - 25\text{ mmHg} = 730\text{ mmHg}

Isolating

P_{\text{dry gas}}

gives the pressure exerted purely by the collected oxygen gas.

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water

Question 468Question →

A block and tackle system consisting of $5$ pulleys is used to raise a load of $200\text{ N}$ through a vertical height of $4\text{ m}$ . If the efficiency of the system is $80\%$ , what is the work done against friction, in joules, during the lifting process?

Show answer & explanation

Answer: 200

Answer

The work done against friction during the lifting process is

200\text{ J}

.

The useful work output is achieved by lifting the

200\text{ N}

load through a vertical height of

4\text{ m}

, yielding

W_{\text{out}} = 200 \times 4 = 800\text{ J}

. Given an efficiency of

80\%

(

0.80

), the total work input required from the effort force is

W_{\text{in}} = \frac{800}{0.80} = 1000\text{ J}

. The energy lost to overcome frictional resistance in the pulleys equals the total work input minus useful work output:

1000\text{ J} - 800\text{ J} = 200\text{ J}

.

Step-by-Step Solution

1

Calculate useful work output

W_{\text{out}} = 800\text{ J}

Useful work output is the energy required to raise the load through the specified height (

W_{\text{out}} = \text{Load} \times \text{height}

).

2

Determine total work input

W_{\text{in}} = 1000\text{ J}

The total work input is calculated from the efficiency formula:

\text{Efficiency} = \frac{W_{\text{out}}}{W_{\text{in}}}

.

3

Compute work done against friction

W_{\text{friction}} = 200\text{ J}

The work lost overcoming friction is the difference between total work input and useful work output (

W_{\text{in}} - W_{\text{out}}

).

Key Concept

Work and Efficiency in Pulley Systems

Estimated Time:1m 30s

Question 469Question →

A capacitor of capacitance $5\text{ }\mu\text{F}$ is charged to a potential difference of $200\text{ V}$ and then disconnected from the power supply. If it is subsequently connected in parallel across an uncharged capacitor of capacitance $15\text{ }\mu\text{F}$ , what is the final common potential difference across the combination in volts?

Show answer & explanation

Answer: 50

Answer

The final common potential difference across the combination is 50 V.

By charge conservation, the total charge

Q = C_1 V_1 = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}

is shared across the two parallel capacitors. The total equivalent capacitance is

C_{\text{total}} = C_1 + C_2 = 20\text{ }\mu\text{F}

. Therefore, the final potential difference is

V = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}

.

Step-by-Step Solution

1

Calculate the initial electric charge (

Q

) stored on the charged capacitor

Q = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}

Before connection, all charge is stored solely on the

5\text{ }\mu\text{F}

capacitor.

2

Calculate the total equivalent capacitance (

C_{\text{total}}

) of the parallel network

C_{\text{total}} = 5\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly (

C_{\text{total}} = C_1 + C_2

).

3

Apply the law of conservation of charge to find the final common voltage (

V

)

V = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}

The total charge remains conserved and redistributes across the total combined capacitance.

Key Concept

Charge Redistribution and Conservation in Parallel Capacitors

Question 470Question →

An enterprise presents the following financial balances at the end of its trading year:

- Premises and Machinery: $\text{₦}1,850,000$
- Stock (Inventory): $\text{₦}410,000$
- Trade Debtors: $\text{₦}240,000$
- Cash at Bank: $\text{₦}150,000$
- Trade Creditors: $\text{₦}270,000$
- Accrued Expenses: $\text{₦}130,000$
- Long-term Loan: $\text{₦}600,000$

What is the Capital Employed of the enterprise in Naira ( $\text{₦}$ )?

Show answer & explanation

Answer: 2250000

Answer

The Capital Employed of the enterprise is

\text{₦}2,250,000

.

Capital Employed represents the total resources employed in the business operations. It is computed as Fixed Assets plus Working Capital. First, calculate Current Assets (Stock

\text{₦}410,000

+ Debtors

\text{₦}240,000

+ Bank

\text{₦}150,000

=

\text{₦}800,000

) and Current Liabilities (Creditors

\text{₦}270,000

+ Accrued Expenses

\text{₦}130,000

=

\text{₦}400,000

). Working Capital is

\text{₦}800,000 - \text{₦}400,000 = \text{₦}400,000

. Adding Working Capital to Fixed Assets (Premises and Machinery

\text{₦}1,850,000

) gives a Capital Employed of

\text{₦}2,250,000

.

Step-by-Step Solution

1

Determine total Current Assets

Current Assets =

\text{₦}410,000 + \text{₦}240,000 + \text{₦}150,000 = \text{₦}800,000

Current assets consist of short-term liquid assets including stock, debtors, and cash at bank.

2

Determine total Current Liabilities

Current Liabilities =

\text{₦}270,000 + \text{₦}130,000 = \text{₦}400,000

Current liabilities consist of short-term obligations payable within a year, including trade creditors and accrued expenses.

3

Calculate Working Capital

Working Capital =

\text{₦}800,000 - \text{₦}400,000 = \text{₦}400,000

Working capital is the net operational buffer calculated as Current Assets minus Current Liabilities.

4

Compute Capital Employed

Capital Employed =

\text{₦}1,850,000 + \text{₦}400,000 = \text{₦}2,250,000

Capital Employed represents the total long-term assets and funds financing the business, calculated as Fixed Assets plus Working Capital (or Total Assets minus Current Liabilities).

Key Concept

Capital Employed represents the total funds actively utilized in running a business. It can be computed either as Fixed Assets + Working Capital or Total Assets - Current Liabilities.

Question 471Question →

A $14.3\text{ g}$ sample of washing soda crystals, hydrated sodium trioxocarbonate(IV) ( $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ ), is completely dissolved in water. Calculate the total number of moles of oxygen atoms present in this sample. [Relative atomic masses: $\text{Na} = 23$ , $\text{C} = 12$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 0.65

Answer

The total number of moles of oxygen atoms present in the sample is 0.65 mol.

The molar mass of

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

is

286\text{ g/mol}

. A mass of

14.3\text{ g}

corresponds to

\frac{14.3}{286} = 0.05\text{ mol}

of the compound. Since each formula unit contains 13 oxygen atoms (3 from the trioxocarbonate anion and 10 from the water of crystallization), the total moles of oxygen atoms present is

0.05 \times 13 = 0.65\text{ mol}

.

Step-by-Step Solution

1

Calculate the molar mass of

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

.

Molar mass =

286\text{ g/mol}

.

Summing the atomic masses of all constituent atoms, including the 10 water molecules of crystallization:

(2 \times 23) + 12 + (3 \times 16) + 10 \times (18) = 286\text{ g/mol}

.

2

Calculate the number of moles of

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

present in the sample.

Moles of compound =

0.05\text{ mol}

.

Dividing the given sample mass by its molar mass:

\frac{14.3\text{ g}}{286\text{ g/mol}} = 0.05\text{ mol}

.

3

Determine the stoichiometric ratio of oxygen atoms per mole of hydrated compound.

13 moles of O atoms per 1 mole of

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

.

Each formula unit contains 3 oxygen atoms from the trioxocarbonate group and 10 oxygen atoms from the ten water of crystallization molecules (

3 + 10 = 13

).

4

Multiply the moles of compound by the number of oxygen atoms per formula unit.

Total moles of O atoms =

0.65\text{ mol}

.

Calculating total oxygen moles:

0.05\text{ mol} \times 13 = 0.65\text{ mol}

.

Key Concept

The Mole Concept, Molar Mass, and Stoichiometric Ratios in Hydrated Salts

Question 472Question →

A galvanometer has an internal resistance of $20\text{ }\Omega$ and produces a full-scale deflection for a current of $15\text{ mA}$ . What resistance of the multiplier, in ohms ( $\Omega$ ), is required to convert this galvanometer into a voltmeter capable of measuring a maximum potential difference of $30\text{ V}$ ?

Show answer & explanation

Answer: 1980

Answer

The required multiplier resistance is

1980\text{ }\Omega

.

To convert a galvanometer into a voltmeter, a multiplier resistor

R_m

is connected in series. The maximum potential difference

V

across the combination is given by

V = I_g(R_g + R_m)

. Substituting

V = 30\text{ V}

,

I_g = 0.015\text{ A}

, and

R_g = 20\text{ }\Omega

yields

R_m = 1980\text{ }\Omega

.

Step-by-Step Solution

1

Convert current from milliamperes to amperes.

I_g = 15\text{ mA} = 0.015\text{ A}

.

Standard electrical formulas require current in amperes.

2

Apply the relationship between voltage range, galvanometer current, internal resistance, and multiplier resistance.

V = I_g(R_g + R_m)

, where

V = 30\text{ V}

,

I_g = 0.015\text{ A}

, and

R_g = 20\text{ }\Omega

.

A multiplier resistor is connected in series with the galvanometer so that the total potential difference is distributed across both components.

3

Rearrange the equation and compute

R_m

.

R_m = \frac{30}{0.015} - 20 = 2000 - 20 = 1980\text{ }\Omega

.

Subtracting the internal resistance of the galvanometer from the total required resistance yields the necessary multiplier resistance.

Key Concept

Galvanometer conversion to a voltmeter using a series multiplier resistor

Estimated Time:1m 30s

Question 473Question →

A resistance thermometer has a resistance of $5.0\,\Omega$ at $0^\circ\text{C}$ and $5.8\,\Omega$ at $40^\circ\text{C}$ . What is the temperature coefficient of resistance of the material in $\text{K}^{-1}$ ?

Show answer & explanation

Answer: 0.004

Answer

The temperature coefficient of resistance of the material is

0.004\,\text{K}^{-1}

(or

4.0 \times 10^{-3}\,\text{K}^{-1}

).

The temperature coefficient of resistance is calculated using

\alpha = \frac{R_T - R_0}{R_0 \Delta T}

. Substituting

R_0 = 5.0\,\Omega

,

R_T = 5.8\,\Omega

, and

\Delta T = 40\,\text{K}

yields

\alpha = \frac{0.8}{200} = 0.004\,\text{K}^{-1}

.

Step-by-Step Solution

1

Identify the relationship between resistance and temperature.

R_T = R_0(1 + \alpha \Delta T)

, with

R_0 = 5.0\,\Omega

,

R_T = 5.8\,\Omega

, and

\Delta T = 40^\circ\text{C}

.

The resistance of metallic conductors varies linearly with temperature for moderate temperature changes.

2

Rearrange the expression to isolate the temperature coefficient

\alpha

.

\alpha = \frac{R_T - R_0}{R_0 \Delta T}

Isolating

\alpha

allows direct computation from the given resistance values and temperature interval.

3

Calculate the numerical value of

\alpha

.

\alpha = \frac{5.8 - 5.0}{5.0 \times 40} = \frac{0.8}{200} = 0.004\,\text{K}^{-1}

Dividing the change in resistance by the product of initial resistance and temperature change gives the fractional resistance change per unit temperature change.

Key Concept

Temperature dependence of electrical resistance and temperature coefficient of resistance.

Question 474Question →

A standard solution is prepared by dissolving $1.575\text{ g}$ of hydrated ethanedioic acid ( $\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}$ ) in distilled water to make $250.0\text{ cm}^3$ of solution. A $25.0\text{ cm}^3$ sample of this acid solution requires $25.0\text{ cm}^3$ of a $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the integer value of $x$ in the formula of the hydrated acid? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Show answer & explanation

Answer: 2

Answer

The integer value of

x

is 2.

The value of

x

is calculated as

2

. Based on the reaction stoichiometry,

25.0\text{ cm}^3

of

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

\text{NaOH}

contains

0.0025\text{ mol}

of base, which neutralizes

0.00125\text{ mol}

of the diprotic acid in the

25.0\text{ cm}^3

sample. The entire

250.0\text{ cm}^3

solution therefore contains

0.0125\text{ mol}

of acid. Dividing the mass (

1.575\text{ g}

) by

0.0125\text{ mol}

gives a molar mass of

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

for

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}

. Subtracting the molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4

(

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

) gives

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

for water, which corresponds to

x = 36 / 18 = 2

.

Step-by-Step Solution

1

Determine the mole ratio from the balanced chemical neutralization equation.

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + (x+2)\text{H}_2\text{O}

. The stoichiometric ratio of acid to base is

1 : 2

.

Ethanedioic acid is a diprotic acid requiring two moles of hydroxide ions for complete neutralization per mole of acid.

2

Calculate the amount in moles of sodium hydroxide solution used in the titration.

\text{Moles of NaOH} = 0.100\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0025\text{ mol}

.

Number of moles is equal to molar concentration multiplied by volume in cubic decimeters.

3

Calculate the total moles of hydrated acid present in the

250.0\text{ cm}^3

volumetric flask.

\text{Moles in } 25.0\text{ cm}^3 \text{ aliquot} = \frac{0.0025}{2} = 0.00125\text{ mol}

. Total moles in

250.0\text{ cm}^3 = 0.00125 \times \frac{250.0}{25.0} = 0.0125\text{ mol}

.

Using the stoichiometric ratio (

n_a/n_b = 1/2

) and scaling up from the

25.0\text{ cm}^3

aliquot to the full

250.0\text{ cm}^3

solution volume.

4

Compute the molar mass of the hydrated acid and solve for

x

.

\text{Molar mass} = \frac{1.575\text{ g}}{0.0125\text{ mol}} = 126\text{ g mol}^{-1}

. Molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4 = 2(1) + 2(12) + 4(16) = 90\text{ g mol}^{-1}

. Mass of

x\text{H}_2\text{O} = 126 - 90 = 36\text{ g mol}^{-1}

. Thus,

x = \frac{36}{18} = 2

.

Subtracting the molar mass of the anhydrous acid from the total molar mass gives the mass of the water of crystallization, which is divided by the molar mass of water (

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

) to find

x

.

Key Concept

Volumetric Analysis and Water of Crystallization Determination

Question 475Question →

Calculate the mass, in grams, of nitrogen contained in a $16.4\text{ g}$ sample of pure calcium trioxonitrate(V), $\text{Ca(NO}_3)_2$ . $[\text{Ca} = 40, \text{N} = 14, \text{O} = 16]$

Show answer & explanation

Answer: 2.8

Answer

The mass of nitrogen in the sample is

2.8\text{ g}

.

The molar mass of

\text{Ca(NO}_3)_2

is calculated as

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

. A

16.4\text{ g}

sample corresponds to

\frac{16.4}{164} = 0.1\text{ mol}

of

\text{Ca(NO}_3)_2

. Because each formula unit contains 2 nitrogen atoms,

0.1\text{ mol}

of compound yields

0.2\text{ mol}

of nitrogen. Multiplying

0.2\text{ mol}

by the molar mass of atomic nitrogen (

14\text{ g/mol}

) gives

2.8\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of calcium trioxonitrate(V),

\text{Ca(NO}_3)_2

164\text{ g/mol}

Sum the relative atomic masses of all atoms present:

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

.

2

Calculate the number of moles of

\text{Ca(NO}_3)_2

present in

16.4\text{ g}

0.1\text{ mol}

Use the formula

\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.4\text{ g}}{164\text{ g/mol}} = 0.1\text{ mol}

.

3

Determine the number of moles of nitrogen atoms in

0.1\text{ mol}

of

\text{Ca(NO}_3)_2

0.2\text{ mol}

of N atoms

Each formula unit of

\text{Ca(NO}_3)_2

contains 2 nitrogen atoms.

4

Calculate the mass of the nitrogen atoms

2.8\text{ g}

Multiply the moles of nitrogen by its atomic mass:

0.2\text{ mol} \times 14\text{ g/mol} = 2.8\text{ g}

.

Key Concept

Mole Concept and Mass Composition of Compounds

Estimated Time:1m 30s

Question 476Question →

A progressive wave traveling along a stretched string has a frequency of $250\text{ Hz}$ and a speed of $300\text{ m/s}$ . What is the minimum distance, in meters, between two points on the string that differ in phase by $\frac{\pi}{3}\text{ rad}$ ?

Show answer & explanation

Answer: 0.2

Answer

The minimum distance between the two points is

0.2\text{ m}

.

The wavelength is found using

\lambda = \frac{v}{f} = \frac{300}{250} = 1.2\text{ m}

. Substituting

\lambda = 1.2\text{ m}

and

\Delta \phi = \frac{\pi}{3}\text{ rad}

into the phase difference formula

\Delta \phi = \frac{2\pi \Delta x}{\lambda}

gives

\Delta x = \frac{(\pi / 3) \times 1.2}{2\pi} = 0.2\text{ m}

.

Step-by-Step Solution

1

Calculate the wavelength ($\lambda$) from wave speed ($v$) and frequency ($f$)

$\lambda = \frac{300}{250} = 1.2\text{ m}$

The fundamental wave equation relates wave speed, frequency, and wavelength as $v = f\lambda$.

2

Apply the phase difference formula to solve for spatial separation ($\Delta x$)

$\Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\pi / 3) \cdot 1.2}{2\pi} = 0.2\text{ m}$

A full cycle of $2\pi\text{ radians}$ corresponds to a spatial displacement of one wavelength ($\lambda$).

Key Concept

Relationship between phase difference and spatial displacement

Question 477Question →

A uniform horizontal wooden beam $AB$ of length $4.0\text{ m}$ and mass $8.0\text{ kg}$ is supported on a pivot placed $1.0\text{ m}$ from end $A$ . A mass of $5.0\text{ kg}$ is suspended from end $B$ . What is the magnitude of the downward vertical force $F$ in newtons that must be applied at end $A$ to keep the beam in horizontal equilibrium? (Take $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 230

Answer

The magnitude of the downward force required at end A to maintain horizontal equilibrium is 230 N.

Taking moments about the pivot, the downward force F at end A produces an anticlockwise moment of F × 1.0 m. This balances the clockwise moments produced by the weight of the beam (80 N × 1.0 m) and the load at end B (50 N × 3.0 m). Equating anticlockwise and clockwise moments gives F × 1.0 = 80 + 150 = 230 N.

Step-by-Step Solution

1

Determine the weights and distance of each force from the pivot point

Weight of beam = 80 N acting at 1.0 m right of pivot; weight at B = 50 N acting at 3.0 m right of pivot; force F acts at 1.0 m left of pivot.

The center of gravity of a uniform 4.0 m beam is at its midpoint (2.0 m from end A).

2

Set up the moment equilibrium equation about the pivot

F × 1.0 = (80 × 1.0) + (50 × 3.0)

For static rotational equilibrium, total anticlockwise moments equal total clockwise moments.

3

Solve for the force magnitude F

F = 230 N

Summing clockwise moments yields 80 + 150 = 230 N m, which divided by 1.0 m gives F = 230 N.

Key Concept

Principle of Moments and Static Equilibrium

Estimated Time:1m 30s

Question 478Question →

A potential difference of $16\,\text{V}$ is applied across a uniform conductor of length $4.0\,\text{m}$ and cross-sectional area $1.5 \times 10^{-6}\,\text{m}^2$ . If the resistivity of the conductor material is $3.0 \times 10^{-7}\,\Omega\cdot\text{m}$ , what is the electric current, in amperes, flowing through the conductor?

Show answer & explanation

Answer: 20

Answer

The electric current flowing through the conductor is

20\,\text{A}

.

The electrical resistance of the wire is first determined using the formula

R = \frac{\rho L}{A} = \frac{(3.0 \times 10^{-7})(4.0)}{1.5 \times 10^{-6}} = 0.8\,\Omega

. Then, by applying Ohm's law (

I = \frac{V}{R}

), the current is computed as

I = \frac{16}{0.8} = 20\,\text{A}

.

Step-by-Step Solution

1

Calculate the electrical resistance of the conductor from its physical dimensions and resistivity.

R = 0.8\,\Omega

Substitute

\rho = 3.0 \times 10^{-7}\,\Omega\cdot\text{m}

,

L = 4.0\,\text{m}

, and

A = 1.5 \times 10^{-6}\,\text{m}^2

into

R = \frac{\rho L}{A}

.

2

Apply Ohm's law to calculate the current flowing through the conductor.

I = 20\,\text{A}

Substitute potential difference

V = 16\,\text{V}

and calculated resistance

R = 0.8\,\Omega

into

I = \frac{V}{R}

.

Key Concept

Relationship between resistivity, resistance, potential difference, and electric current

Question 479Question →

A transverse progressive wave traveling along a stretched string is represented by the mathematical wave equation $y = 0.05 \sin(160\pi t - 8\pi x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the speed of propagation of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 20

Answer

The speed of propagation of the wave is

20\text{ m/s}

.

Comparing the given equation

y = 0.05 \sin(160\pi t - 8\pi x)

with the standard wave equation

y = A \sin(\omega t - kx)

gives the angular frequency

\omega = 160\pi\text{ rad/s}

and wave number

k = 8\pi\text{ rad/m}

. Substituting these into

v = \frac{\omega}{k}

yields

v = \frac{160\pi}{8\pi} = 20\text{ m/s}

.

Step-by-Step Solution

1

Compare the given wave equation with the standard progressive wave equation.

The standard form is

y = A \sin(\omega t - kx)

. Comparing parameters yields

\omega = 160\pi\text{ rad/s}

and

k = 8\pi\text{ rad/m}

.

Matching coefficients allows direct extraction of angular frequency and wave number.

2

Calculate the wave speed using the relation between angular frequency and wave number.

v = \frac{\omega}{k} = \frac{160\pi\text{ rad/s}}{8\pi\text{ rad/m}} = 20\text{ m/s}

.

Wave speed is defined as the ratio of angular frequency to wave number (

v = \lambda f = \frac{\omega}{k}

).

Key Concept

Wave Equation Parameter Extraction and Wave Speed Calculation

Question 480Question →

An aqueous solution of a weak monoacidic base, $\text{BOH}$ , has a concentration of $0.20\text{ mol dm}^{-3}$ and a base dissociation constant $K_b = 5.0 \times 10^{-6}\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Calculate the pH of this solution.

Show answer & explanation

Answer: 11

Answer

The pH of the weak monoacidic base solution is 11.0.

For a weak base, partial equilibrium ionization gives

[\text{OH}^-] = \sqrt{K_b \times C} = \sqrt{5.0 \times 10^{-6} \times 0.20} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The pOH is

-\log_{10}(1.0 \times 10^{-3}) = 3.0

. Using

\text{pH} = 14.0 - \text{pOH}

, the pH of the basic solution is

14.0 - 3.0 = 11.0

.

Step-by-Step Solution

1

Formulate the dissociation equilibrium equation for the weak base BOH.

The ionization is

\text{BOH}_{(aq)} \rightleftharpoons \text{B}^+_{(aq)} + \text{OH}^-_{(aq)}

, giving

K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}

.

Weak bases ionize incompletely in aqueous media.

2

Calculate the equilibrium hydroxide ion concentration [OH⁻].

[\text{OH}^-] = \sqrt{K_b \times C} = \sqrt{(5.0 \times 10^{-6})(0.20)} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Since the base is weak and degree of ionization is small,

[\text{BOH}] \approx 0.20\text{ mol dm}^{-3}

and

[\text{B}^+] = [\text{OH}^-]

.

3

Determine the pOH of the solution.

\text{pOH} = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

pOH is defined as

-\log_{10}[\text{OH}^-]

.

4

Calculate the pH using the relationship between pH and pOH at 25°C.

\text{pH} = 14.0 - \text{pOH} = 14.0 - 3.0 = 11.0

.

At

25^\circ\text{C}

,

\text{pH} + \text{pOH} = 14.0

.

Key Concept

Weak base dissociation equilibrium and pH determination

Answer

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