Question

Calculate the boiling point of a solution containing 0.61 g of benzoic acid in 50 g of carbon disulphide assuming 84 per dimerization of the acid. The boiling point and $K_b$ of $C{S}_2$ are ${46.2}^{\circ }$C and 2.3 K kg mol${}^{-1},$respectively.

• A. ${46.33}^{\circ }$C
• B. ${90.4}^{\circ }$C
• C. ${12.3}^{\circ }$C
• D. ${62.21}^{\circ }$C

Answer 1

The correct option is A ${46.33}^{\circ }$C

Elevation of boiling-point describes that the boiling point of a liquid (a solvent) will be higher when another compound is added to it, meaning that a solution has a higher boiling point than a pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water.

Elevation of boiling point ( $\Delta T_b$ ) is the difference between the boiling points of the solution and the pure solvent:

$\Delta T_b=T_b-{T_b}^0$

Elevation of boiling point can be calculated as

$\Delta T_b=i.m.K_b$

Here, m is the concentration of the solute expressed in molality and $K_b$ is the molal boiling point elevation constant of the solvent.

Degree of association will is $84\%$.

Hence, $x=0.84$

For benzoic acid in carbon disulphide,

$2C_6{H}_{5}COOH\leftrightharpoons {{(C}_6{H}_{5}COOH)}_2$

If $x$ represents the degree of association of the solute, then we would have $(1–x)$ mol of benzoic acid left in unassociated form and correspondingly 2 x as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is

$i=1-\frac{x}{2}=0.58$

No. of moles of benzoic acid is $\frac{0.61}{122}=0.05$

Molality is $m=\frac{0.005}{0.05}=0.1$

The elevation of boiling point is $i.m.K_b=2.3\times 0.58\times 0.1=0.1334$

$T_b=46.2+0.1334={46.33}^{0}C$