Question

A certain substance 'A' tetramerises in water to the extent of 80%. A solution of $2.5g$ of A in $100g$ of water lowers the freezing point by ${0.3}^{\circ }C.$ The molar mass of A in (g/mol) is:
(Molal depression constant of A is $1.86Kkgmo{l}^{-1}$)

• A. 122
• B. 31
• C. 244
• D. 62

Answer 1

The correct option is D 62

For a electrolyte undergoing association,
$\beta =\frac{(i-1)n}{(1-n)}$
where,
$\beta =\text{degree of association}$
$i=\text{is van't Hoff factor}$
$n=\text{No. of ions of the electrolyte}$
n = 4 since, the substance is tetramerises.
Thus,
$0.8=\frac{(i-1)4}{(1-4)}$
$-2.4=4i-4$
$4i=1.6$
$i=0.4$

For depression of freezing point of the solution,
$△T_f=i\times K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of solution.
Therefore,
$0.3=0.4\times 1.86\times \frac{w_B\times 1000}{m_B\times w_A}$
where,
$w_B$ is mass of solute
$m_B$ is molar mass of solute
$w_A$ is mass of solvent
$0.3=0.4\times 1.86\times \frac{2.5\times 1000}{m_B\times 100}{m}_B=62$
The molar mass of A is $62gmo{l}^{-1}$