Question

Substance $A$ tetramerises in water to the extent of $80$. A solution of $2.5g$ of $A$ is $100g$ of water lowers the freezing point by ${0.3}^{o}C$. The molar mass of $A$ is ($K_f$ for water $=1.86KKgmo{l}^{-1}$)

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

Answer 1

The correct option is D $62$

Since substance $A$ tetra menses in water so equation can be written as

initially $4A\underset{}{\overset{H_{2}O}{\to }}A4$

$c$ $o$

At any time $t$ $C(1-4\alpha )$ $c\alpha$.

unit half factor $=\frac{c(1-4\alpha )+c\alpha }{c}$

$1=1-3\alpha$.

$4\alpha =80\%\Rightarrow \alpha =20$.

$i=1-3\times 0.2=0.4$

Molar man of $A=M_A$

morality $=\frac{2.5}{MA}/100g\times {10}^{-3}kg$.

$=\frac{25}{MA}$

$△T_f=ikgm$

$0.3=0.4\times 1.86\times \frac{25}{MA}\Rightarrow MA=\frac{25\times 1.86\times 0.4}{0.3}$

$MA=62g$