# Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of benzene. The freezing point of benzene is 5.5oC, and the freezing point of the mixture is 2.8oC. What is the molal freezing point constant for benzene?

## Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of benzene. The freezing point of benzene is 5.5oC, and the freezing point of the mixture is 2.8oC. What is the molal freezing point constant for benzene?

May 11, 2016

Your experimental molal freezing point depression constant for benzene is $\text{4.3 °C·kg·mol"^"-1}$.

#### Explanation:

The formula for calculating freezing point depression is

color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_fm color(white)(a/a)|)))" "

where

ΔT_"f" is the decrease in freezing point

${K}_{\text{f}}$ is the molal freezing point depression constant

$\text{m}$ is the molality of the solution.

We can rearrange the above formula to get

K_"f" = (ΔT_"f")/m

ΔT_"f" = T_"f"^° -T_"f" = "5.5 °C – 2.8 °C" = "2.7 °C"

${\text{Moles of C"_10"H"_8 = 1.60 color(red)(cancel(color(black)("g C"_10"H"_8))) × ("1 mol C"_10"H"_8)/(128.17 color(red)(cancel(color(black)("g C"_10"H"_8)))) = "0.012 48 mol C"_10"H}}_{8}$

$\text{Molality" = "moles of solute"/"kilograms of solvent" = "0.012 48 mol"/"0.0200 kg" = "0.6242 mol/kg}$

${K}_{\text{f" = (ΔT_"f")/m = "2.7 °C"/"0.6242 mol/kg" = "4.3 °C·kg·mol"^"-1}}$

The actual value of ${K}_{\text{f}}$ for benzene is $\text{5.1 °C·kg·mol"^"-1}$.

Are you sure that the observed freezing point wasn't 2.3 °C?