Calculate the freezing point of a 0.05 m aqueous solution of a non-electrolyte at 298 K. ( K f = 1.86 K kg mol − 1 )

Question

Calculate the freezing point of a 0.05 m aqueous solution of a non-electrolyte at 298 K. ( K f ​ = 1.86 K kg mol − 1 )

TestHub · Accepted Answer

For a non-electrolyte, the van 't Hoff factor $i = 1$ .

The depression in freezing point is given by:

$Δ T_{f} = i \times K_{f} \times m$

Substituting the given values:

$Δ T_{f} = 1 \times 1.86 \times 0.05 = 0.093$ K

The freezing point of pure water is $0^{\circ} C$ .

The freezing point of the solution is:

$T_{f} = T_{f}^{\circ} - Δ T_{f} = 0 - 0.093 = - 0.093^{\circ} C$

Answer: $- 0.093^{\circ} C$

Chemistry - Liquid Solution Question with Solution | TestHub

Options:

Doubts & Discussion