Question:
How much amount of $\mathrm{NaCl}$ should be added to $600 \mathrm{~g}$ of water $(\rho=1.00 \mathrm{~g} / \mathrm{mL})$ to decrease the freezing point of water to $-0.2^{\circ} \mathrm{C}$ ? _________________ .
(The freezing point depression constant for water $=2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
Solution:
(1.75)
$\Delta \mathrm{T}_{\mathrm{f}}=i k_{\mathrm{f}} m$
$0.2=2 \times 2 \times \frac{w}{58.5} \times \frac{1000}{600}$
$w=\frac{0.2 \times 58.5 \times 600}{1000 \times 4}=\frac{1.2 \times 58.5}{40}=1.75 \mathrm{~g}$