Question

Equal volumes of ethylene glycol (molar mass $= 62$) and water (molar mass $=18$ ) are mixed. The depression in freezing point of water is (given $K _{ f }$ of water $=$ $1.86 \,K$ mor $^{-1} kg$ and specific gravity of ethylene glycol is $1.11$ )

• A. 0.0033
• B. 0.033
• C. 0.33
• D. 33 .3

Answer 1

Answer: (D)

$M=D \times V$
$M$ (Ethylene glycol) $=1.11 \times v$
$M$ (water) $=1.00 \times v$
$n$ (Ethylene glycol) $=\frac{1.11 v}{62} mol$
$\Delta T_{f}=K_{f} \times m=1.86 \times \frac{1.11 v}{62} \times \frac{1000}{1.00 v}$
$=\frac{1.86 \times 1.11 \times 1000}{62}=33.3$