1.00 g of a non-electrolyte solute (molar mass 250g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by :

Question

1.00 g of a non-electrolyte solute (molar mass 250g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by : (A) 0.4 K (B) 0.8 K (C) 0.12 K (D) 0.24 K

Tardigrade Admin · Accepted Answer

Δ T f - text molality × K f Δ T f =((1 × 1000/250 × 51.2)) × 5.12 Δ T f =0.4 K

Q. 1.00 g of a non-electrolyte solute (molar mass 250g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by :

0.4 K

0.8 K

0.12 K

0.24 K

$Δ T_{f} - molality \times K_{f}$

$Δ T_{f} = (\frac{1 \times 1000}{250 \times 51.2}) \times 5.12$

$Δ T_{f} = 0.4 K$

Q. 1.00 g of a non-electrolyte solute (molar mass 250g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by :

0.4 K

0.8 K

0.12 K

0.24 K

ΔTf​− molality ×Kf​ ΔTf​=(250×51.21×1000​)×5.12 ΔTf​=0.4K

$Δ T_{f} - molality \times K_{f}$

$Δ T_{f} = (\frac{1 \times 1000}{250 \times 51.2}) \times 5.12$

$Δ T_{f} = 0.4 K$