How much amount of NaCl should be added to 600 g of water

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)

$\Delta \mathrm{T}_{\mathrm{f}}=i k_{\mathrm{f}} m$

$0.2=2 \times 2 \times \frac{w}{58.5} \times \frac{1000}{600}$

$w=\frac{0.2 \times 58.5 \times 600}{1000 \times 4}=\frac{1.2 \times 58.5}{40}=1.75 \mathrm{~g}$

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