Elevation in boiling point of an aqueous solution of a non-electrolyte solute is 2.01° . What is the depression in freezing point of this solution? Kb( H 2 O )=0.52° mol -1 kg Kf( H 2 O )=1.86° mol -1 kg

Question

Elevation in boiling point of an aqueous solution of a non-electrolyte solute is 2.01° . What is the depression in freezing point of this solution? Kb( H 2 O )=0.52° mol -1 kg Kf( H 2 O )=1.86° mol -1 kg (A) 3.72° (B) 0.52° (C) 7.17° (D) 0.93°

Tardigrade Admin · Accepted Answer

Δ Tb= Molality × Kb Δ Tb =m Kb Δ Tf =m Kf ∴ (Δ Tf/Δ Tb) =(Kf/Kb) Δ Tf =Δ Tb (Kf/Kb) =2.01 × (1.86/0.52)=7.17°

Q. Elevation in boiling point of an aqueous solution of a non-electrolyte solute is $2.0 1^{\circ} .$ What is the depression in freezing point of this solution?
$K_{b} (H_{2} O) = 0.5 2^{\circ} m o l^{- 1} k g$
$K_{f} (H_{2} O) = 1.8 6^{\circ} m o l^{- 1} k g$

$3.7 2^{\circ}$

$0.5 2^{\circ}$

$7.1 7^{\circ}$

$0.9 3^{\circ}$

$Δ T_{b} =$ Molality $\times K_{b}$

$Δ T_{b} = m K_{b}$

$Δ T_{f} = m K_{f}$

$∴ \frac{Δ T _{f}}{Δ T _{b}} = \frac{K _{f}}{K _{b}}$

$Δ T_{f} = Δ T_{b} \frac{K _{f}}{K _{b}}$

$= 2.01 \times \frac{1.86}{0.52} = 7.1 7^{\circ}$

Q. Elevation in boiling point of an aqueous solution of a non-electrolyte solute is 2.01∘. What is the depression in freezing point of this solution? Kb​(H2​O)=0.52∘mol−1kg Kf​(H2​O)=1.86∘mol−1kg

3.72∘

0.52∘

7.17∘

0.93∘