Question

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]

पर्याय

• 271

• 360

• 300

• 400

Answer 1

40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is 271 K.

Explanation:

molality = `((40/180)"mol")/(0.2 " kg") = (10/9)` molal

⇒ Δ T_f = T_f - T_f' = `1.86 xx 10/9`

⇒ T_f' = 273.15 - `1.86 xx 10/9`

= 271.08 K ≈ 271 K