Question

For an aqueous solution, freezing point is −0.186°C. Elevation of the boiling point of the same solution (K_f = 1.86° mol⁻¹ kg and K_b = 0.512° mol⁻¹ kg) is ______.

पर्याय

• 0.186°

• 0.0512°

• 1.86°

• 5.12°

Answer 1

For an aqueous solution, freezing point is −0.186°C. Elevation of the boiling point of the same solution (K_f = 1.86° mol⁻¹ kg and K_b = 0.512° mol⁻¹ kg) is 0.0512°.

Explanation:

Given: ΔT_f​ = 0.186°CK_f = 1.86° mol⁻¹ kg

K_b = 0.512° mol⁻¹ kg

We know that

ΔT_f ​= K_f​ . m

`m = (Delta T_f)/K_f`

= `0.186/1.86`

m = 0.1 mol/kg

ΔT_b​ = K_b m

= 0.512 × 0.1

= 0.0512°C