Question

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0.694 g of biphenyl (C12H10) in 25.0 g of cyclohexane?

Answer 1

Answer:

The freezing point will be $2.9^{0}C$

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

$Depressioninfreezing point=K_{f}Xmolality$

Where

Kf= $20\frac{^{0}C}{m}$

the molality is calculated as:

$molality=\frac{moles_{solute}}{mass_{solvent}}$

$moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol$

$massofcyclohexane=25g=0.025Kg$

$molarity=\frac{0.0045}{0.025}=0.18m$

Depression in freezing point = $20X0.18=3.6^{0}C$

The new freezing point = $6.5^{0}C-3.6^{0}C=2.9^{0}C$