Question

How much amount of NaCl should be added to 500 g of water ( $\rho =1.00g/mL$ ) to decrease the freezing point of water to -0.3 $^\circ $ C?
(The freezing point depression constant for water $=2 \, Kkgmol^{- 1}$ )

• A. 2.19 g
• B. 1.88 g
• C. 1.96 g
• D. 1.085 g

Answer 1

Answer: (A)

$\Delta T_{f}=0.3$
$\Delta T_{f}=iK_{f}m$
$i=2$ for NaCl
$0.3=2\times 2\times m$
$m=\frac{w}{58.5}\times \frac{1000}{500}$
$0.3=2\times 2\times \frac{w}{58.5}\times \frac{1000}{500}$
$w=\frac{0.3 \times 58.5 \times 500}{4 \times 1000}=2.19g$