Question

Molality of pure liquid benzene $(C_6H_6)$ if its density $d = 2$ gram/ml is -

• A. 2 m
• B. 0.78 m
• C. 12.8 m
• D. 50 lution of $CH_3COOH$ is 0.1, Select incorrect state

Answer 1

Answer: (C)

12.8 m

Answer 2

To calculate the molality of a pure liquid, we consider the substance itself as both the solute and the solvent.

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

$$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$$

For a pure liquid like benzene ($C_6H_6$), let's consider a sample of 1 kilogram (1000 grams) of the liquid. In this context, this 1 kg of benzene acts as the 'solvent'. The 'solute' would then be the moles of benzene present in this 1 kg.

Step 1: Determine the molar mass of benzene ($C_6H_6$).

The atomic mass of Carbon (C) is approximately 12.01 g/mol.

The atomic mass of Hydrogen (H) is approximately 1.008 g/mol.

Molar mass of $C_6H_6 = (6 \times \text{Atomic mass of C}) + (6 \times \text{Atomic mass of H})$

Molar mass of $C_6H_6 = (6 \times 12.01 \text{ g/mol}) + (6 \times 1.008 \text{ g/mol})$

Molar mass of $C_6H_6 = 72.06 \text{ g/mol} + 6.048 \text{ g/mol} = 78.108 \text{ g/mol}$

For practical purposes in competitive exams, often rounded values are used:

Molar mass of $C_6H_6 \approx (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \text{ g/mol}$.

Step 2: Calculate the number of moles of benzene in 1 kg (1000 g).

Moles of benzene = $\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}$

Moles of benzene = $\frac{1000 \text{ g}}{78 \text{ g/mol}} \approx 12.82 \text{ mol}$

Step 3: Calculate the molality.

Since we considered 1 kg of benzene as the 'solvent', and the moles of benzene in it are 12.82 mol, the molality is:

Molality = $\frac{12.82 \text{ mol}}{1 \text{ kg}} = 12.82 \text{ m}$

The density of benzene (d = 2 gram/ml) given in the question is a distractor. For the molality of a pure liquid, the density is not required as it cancels out if we consider any arbitrary volume, or it is not used if we directly consider 1 kg of the substance.

For example, if we consider a volume $V$ (in mL) of pure benzene:

Mass of benzene = $V \times d$ (in grams)

Moles of benzene = $\frac{V \times d}{M}$

Mass of solvent (in kg) = $\frac{V \times d}{1000}$

Molality = $\frac{\text{Moles of benzene}}{\text{Mass of solvent (in kg)}} = \frac{(V \times d) / M}{(V \times d) / 1000} = \frac{V \times d}{M} \times \frac{1000}{V \times d} = \frac{1000}{M}$

As shown, the density ($d$) and volume ($V$) cancel out.

The calculated molality is approximately 12.8 m.