Mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kf is 1.86°C mol-1 kg, depression in f.p t or Δ Tf) is

Question

Mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kf is 1.86°C mol-1 kg, depression in f.p t or Δ Tf) is (A) 0.26°C (B) 1.86°C (C) 0.13°C (D) 7.78°C

Tardigrade Admin · Accepted Answer

Firstly we have to convert mole fraction into molality. Molality = (x textsolute/x textsolvent M textsolvent 1000) = (0.07 × 1000/0.93 × 18) = 4.18 Now, Δ Tf = Kfm = 1.86 × 4.18 = 7.78°C

Q. Mole fraction of a non-electrolyte in aqueous solution is $0.07$ . If $K_{f}$ is $1.8 6^{\circ} C m o l^{- 1} k g$ , depression in f.p t or $Δ T_{f}$ ) is

$0.2 6^{\circ} C$

$1.8 6^{\circ} C$

$0.1 3^{\circ} C$

$7.7 8^{\circ} C$

Firstly we have to convert mole fraction into molality.
Molality $= \frac{x _{solute}}{x _{solvent} M _{solvent} 1000}$
$= \frac{0.07 \times 1000}{0.93 \times 18} = 4.18$
Now, $Δ T_{f} = K_{f} m = 1.86 \times 4.18 = 7.7 8^{\circ} C$

Q. Mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kf​ is 1.86∘Cmol−1kg, depression in f.p t or ΔTf​) is

0.26∘C

1.86∘C

0.13∘C

7.78∘C