# 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?

##### 1 Answer
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?