Question

How many grams of $KCl$ should be added to $0.1\, kg$ of water to lower its freezing point to $-8.0^{\circ} C ?\left[K_{f}\right.$ for water $=1.86^{\circ} C\,kg\, mol ^{-1}$ )

• A. 160.2
• B. 150.2
• C. 140.2
• D. 130.2

Answer 1

Answer: (A)

$\Delta T_{f}=\frac{K_{f} \times w \times 1000}{m \times W}$
Given, $\Delta T_{f}=8^{o} C K_{f}=1.86^{\circ} C\, kg\, mol ^{-1}$
$m=$ mass of solute $=35.5+19=54.5$
$w =1\, kg$
$\therefore 8=\frac{1.86 \times w \times 100}{54.5 \times 1} W =320.5\, g$
$KClK^{+}+C l^{-}(i=2)$ Hence,
$w=\frac{w}{2}=\frac{320.5}{2}=160.2\, g$