Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol−1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol−1] Among the following, the option representing change in the freezing point is

Question

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol−1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol−1] Among the following, the option representing change in the freezing point is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ T f = K f × m =2 × (34.5 × 2/46) =2 × 1.5 =3 Freezing point of ethanol + water mixture =273-3=270

ΔTf​=Kf​×m =2×4634.5×2​ =2×1.5 =3 Freezing point of ethanol + water mixture =273−3=270

$Δ T_{f} = K_{f} \times m$
$= 2 \times \frac{34.5 \times 2}{46}$
$= 2 \times 1.5$
$= 3$
Freezing point of ethanol $+$ water mixture
$= 273 - 3 = 270$