Question

An aqueous solution freezes at −0.372°C. The boiling point of the same solution is ___________.

(K_f = 1.86 K m⁻¹, K_b = 0.512 K m⁻¹)

Options

• 100.186°C

• 100.512°C

• 99.949°C

• 100.1024°C

Answer 1

An aqueous solution freezes at −0.372°C. The boiling point of the same solution is 100.1024°C.

Explanation:

Depression in freezing point = ΔT_f

Molal concentration = m

Molal depression constant = K_f

Since freezing point of water = 0°C

ΔT_f = 0.372°C ............(∵ ΔT_f = 0.372 K)

ΔT_f = K_f × m

∴ m = `(Δ"T"_"f")/("K"_"f")`

= `(0.372  0.372  "K")/(1.86  0.512  "K m"^-1)`

= 0.2 m

Here, ΔT_b = Elevation in boiling point

K_b = molal elevation constant

ΔT_b = K_b × m

∴ ΔT_b = 0.512 K m⁻¹ × 0.2 m = 0.1024 K .....(∵ ΔT_b = 0.1024°C)

Boiling point of solution

= Boiling point of water + ΔT_b

= 100°C + 0.1024°C

= 100.1024°C