substance A tetramerises in water to a extent of 80% . A solution of 2.5g of A in 100g of water lowers the freezing point by 0.3degree C . The molar mass of A is

1) 122

2) 31

3) 244

4) 62

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

​Molality of solution = No. of moles of solute / kg of solvent
                             = (Mass/Molar mass) / kg of solvent  =$\frac{(m/M)}{Mass of solvent}= \frac{(2.5/M)}{100/1000}$

Depression in freezing point  =$\Delta T_f =i K_{f}m$ 
 $\Delta T_f = 0.3 K$
$0.3 =i K_{f}m$

$$\begin{gathered} \alpha =\frac{i -1}{{\displaystyle \frac{i}{m}}-1} \\ or \\ i =\alpha (\frac{1}{m}-1)+1 =0.8(\frac{1}{4}-1) + 1 =0.4 \end{gathered}$$


$$\begin{gathered} 0.3 =0.4 \times 1.86 \times m \\ m=\frac{0.3}{0.4\times 1.86} \\ \frac{0.3}{0.4\times 1.86}=\frac{2.5/M}{0.1} \\ Molar mass\left(M\right) = \frac{2.5\times 0.4\times 1.86}{0.1\times 0.3}=62 g/mol \end{gathered}$$