Question

The freezing point of pure benzene is 278.4 K. When 4.0 g of a solute having molecular weight 100 g mol⁻¹ is added to 200 g of benzene, the freezing point of the solution will be ____________.

(K_f of benzene = 5.12 K kg mol⁻¹)

पर्याय

• 276.236 K

• 278.400 K

• 279.424 K

• 277.376 K

Answer 1

The freezing point of pure benzene is 278.4 K. When 4.0 g of a solute having molecular weight 100 g mol⁻¹ is added to 200 g of benzene, the freezing point of the solution will be 277.376 K.

Explanation:

∆T_f = `(1000  "K"_"f""W"_2)/("M"_2"W"_1)`

= `(1000  "g kg"^-1 xx 5.12  "K kg mol"^-1 xx 4  "g")/(100  "g mol"^-1 xx 200  "g")`

= 1.024 K

∆T_f = `"T"_"f"^0 - "T"_"f"`

∴ ∆T_f = `"T"_"f"^0 - ∆"T"_"f"`

∆T_f = 278.4 K − 1.024 K = 277.376 K