Question

$1.22g$ of an organic acid is separately dissolved in $100g$ of benzene $(K_b=2.6Kkgmo{1}^{-1})and100g$ of acetone $(K_b=1.7Kkgmo{1}^{-1})$. The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17°C$. The increase in boiling point of solution in benzene in $°Cisx\times {10}^{-2}$. The value of $x$ is _____________. (Nearest integer)
(Atomic mass : $C=12.0,H=1.0,O=16.01]$

• A. 13.0
• B. 13.00
• C. 13

Answer 1

Answer: (A), (B), (C)

$K_b\left(benzene\right)=2.6Kkgmo{l}^{-1}$

$K_b\left(acetone\right)=1.7Kkgmo{l}^{-1}$

Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.

$\text{Molality (m)}=\frac{n_{\text{solute}}\times 1000}{W_{\text{solvent}}\text{in gram}}$
$\text{Molality (m)}=\frac{w_{\text{solute}}\times 1000}{M_{\text{solute}}.W_{\text{solvent}}\text{in gram}}$
where,
$w_{\text{solute}}$ is mass of solute
$W_{\text{solvent}}$ is mass of solvent
$M_{\text{solute}}$ is molar mass of solute

In acetone:

$0.17=\frac{1\times 1.7\times 1.22\times 1000}{M_{solute}\times 100}$

$M_{solute}=122g/mol$

In benzene:
Acid will dimerise and considering it as complete association
$i=\frac{2-\alpha }{2}$
$i=\frac{2-1}{2}\text{(Complete association})$
$i=\frac{1}{2}$
$\Delta T_b=\frac{1}{2}\times \frac{1.22}{122}\times 2.6\times \frac{1000}{100}$

$={0.13}^{o}C$

$=13\times {10}^{-2}{}^{o}C$