How much amount of NaCl should be added to 600 g of water ($\rho=1.00$ $\mathrm{g/mL}$) to decrease the freezing point of water to $-0.2^{\circ}\mathrm{C}$? ____. (The freezing point depression constant for water $=2$ $\mathrm{K}$ $\mathrm{kg}$ $\mathrm{mol}^{-1}$)

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Solutions

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

ρ = 1.00

g/mL

) to decrease the freezing point of water to

- 0. 2^{\circ} C

? ____.
(The freezing point depression constant for water

= 2

K

kg

mol^{- 1}

)

Answer: 1.76

Solution

The freezing point is lowered by

Δ T_{f} = 0.2 K

, and sodium chloride dissociates into two ions so its van't Hoff factor is

i = 2

. Using

Δ T_{f} = i K_{f} m

, the required molality is

m = \frac{Δ T _{f}}{i K _{f}} = \frac{0.2}{2 \times 2} = 0.05 mol kg^{- 1}

. The solvent is

600 g = 0.6 kg

of water, so the amount of salt needed is

0.05 \times 0.6 = 0.03 mol

. Multiplying by the molar mass of NaCl,

0.03 \times 58.5 = 1.755 \approx 1.76 g

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Numerical answer

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