Question

When 2.0 g of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point of benzene is raised by 0.88 K. Which of the following may be the solute ?$(K_b\, for \,benzene = 2.53 K kg mol^{-1})$

• A. $CO(NH_2)_2$
• B. $C_6H_{12}O_6$
• C. NaCl
• D. None of these

Answer 1

The Correct Option is A

$Na^+Cl^-$ an ionic solid is not soluble in benzene
$\therefore$ $\Delta T_b = K_b \times m$
$m = \frac{\Delta T_b}{K_b} = \frac{0.88}{2.53} $ = 0.348
$m = \frac{w \times 1000}{M \times W}$
0.348 = $\frac{2 \times 1000}{M \times 90}$
[W = mass of solvent, $w$ = mass of solute]
$M = \frac{2 \times 1000}{90 \times 0.348}$ = 63.86
Molar mass of urea, $CO(NH_2)_2$
= 12 + 16 + (14 + 2) $\times$ 2
= 28 + 32 = 60 g $mol^{-1}$
$\therefore$ Solute is urea (A)