Question

Solute ‘$X$’ dimerises in water to the extent of $80\%. 2.5\,g$ of ‘$X$’ in $100\,g$ of water increases the boiling point by $0.3 ^\circ C$. The molar mass of ‘$X$’ is $[K_b=0.52\,K \,kg \,mol^{-1}]$

• A. 13
• B. 52
• C. 65
• D. 26

Answer 1

Answer: (D)

We are given,

$2 X \rightarrow X _{2}$

Therefore, $i =1-\alpha+\propto / 2=1-0.8+0.4=0.6$

(since $\alpha=0.8$, given $)$

We know, $\Delta T _{ b }= ik _{ b } \times W _{ B } / M \times 1000 /\left( W _{ A }( g )\right)$

Where, $W _{ B }=$ Mass of solute,

$W _{ A }=$ Mass of solvent,

$M =$ Molar mass

Given, $\Delta T _{ b }=0.3$

Therefore, $0.3=0.6 \times 0.52 \times 2.5 / M \times 1000 / 100$

$\Rightarrow M =2 \times 0.52 \times 2.5 \times 10=26 \,g$