40 g of glucose (Molar mass $=180$ ) is mixed with 200 mL of water. The freezing point of solution is ____ K. (Nearest integer) [Given : $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; Density of water $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$; Freezing point of water $=273.15 \mathrm{~K}]$.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Solutions

40 g of glucose (Molar mass

= 180

) is mixed with 200 mL of water. The freezing point of solution is ____ K. (Nearest integer)
[Given :

K_{f} = 1.86 K kg mol^{- 1}

; Density of water

= 1.00 g cm^{- 3}

; Freezing point of water

= 273.15 K]

Answer: 271

Solution

The solution contains

40/180 = 0.222

mol glucose in

200 mL = 0.200 kg

of water, a molality of

0.222/0.200 = 1.11 mol k g^{- 1}

. The freezing-point depression is

Δ T_{f} = K_{f} m = 1.86 \times 1.11 = 2.06 K,

so the freezing point of the solution is

273.15 - 2.06 = 271 K

to the nearest integer.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer

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