Question

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 ${0.3}^{o}C$.The molar mass of $A$ is:
$[kf={1.86}^{o}Cm-1]$

• A. $122$
• B. $31$
• C. $244$
• D. $62$

Answer 1

Answer: (D)

$62$

Mass of solute $=2.5g$

Mass of solution $=(100+2.5)g=102.5g$

As the substance A tetramerises in water to the extent of $80\%$

Degree of association $\left(\alpha \right)=0.8$

Molality of solution (m) $=\frac{\text{No. of moles of solute}}{\text{kg of solvent}}=\frac{\left(\frac{\text{mass of solute}}{\text{molar mass of solute}}\right)}{\text{Kg of solvent}}$

$\Rightarrow m=\frac{\left(\frac{2.5}{M}\right)}{\left(\frac{100}{1000}\right)}=\frac{25}{M}$

Van't hoff constant $\left(i\right)=\alpha (\frac{1}{n}-1)+1$

As the substance A tetramerises, $n=4$

$\therefore i=(0.8\times (\frac{1}{4}-1))+1=0.4$

As we know that, depression in freezing point,

$\Delta T_f=i{k}_f.m$

where,

$\Delta T_f$ = Elevation in freezing point = $0.3℃$

$k_f$ = Freezing point depression constant = $1.86℃m^{-1}$

$m$ = molality of solution = $\frac{25}{M}$

$i$ = Van't hoff constant = $0.4$

$\therefore 0.3=0.4\times 1.86\times \frac{25}{M}$

$\Rightarrow 0.3=\frac{18.6}{M}$

$\Rightarrow M=62g{mole}^{-1}$