# Question accdb

May 31, 2015

The molal cryoscopic constant for benzene will be equal to ${4.3}^{\circ} \text{C/molal}$.

You're dealing with a solution that consists of naphthalene, your solute, and benzene, your solvent. In order to determine the cryoscopic constant of benzene, you'll need to calculate the molality of the solution.

Molality is defined as moles of solute per kilogram of solvent. Use naphthalene's molar mass to determine how many moles you have

1.60cancel("g") * ("1 mole "C_10H_8)/(128.17cancel("g")) = "0.01248 moles"

The molality of the solution will thus be

$b = {n}_{{C}_{10} {H}_{8}} / {m}_{{C}_{6} {H}_{6}} = \text{0.01248 moles"/(20.0 * 10^(-3)"kg") = "0.624 molal}$

The freezing point depression equation looks like this

$\Delta {T}_{\text{f}} = i \cdot {K}_{f} \cdot b$, where

${K}_{f}$ - the cryoscopic constant;
$b$ - the molality of the solution;
$i$ - the van't Hoff factor - the number of ions per individual molecule of solute.
$\Delta {T}_{\text{f}}$ - the freezing point depression - defined as
${T}_{\text{f"^@ - T_"f sol}}$.

Since you're dealing with a non-electrolyte, the van't Hoff factor will be equal to 1.

Since you know the freezing point of benzene and of the solution, you can write

T_"f"^@ - T_"f sol" = K_f * b => K_f = (T_"f"^@ - T_"f sol")/b

K_f = ((5.5 - 2.8)^@"C")/("0.624 molal") = 4.327^@"C/molal"#

Rounded to two sig figs, the number of sig figs you gave for the freezing points, the answer will be

${K}_{f} = \textcolor{g r e e n}{{4.3}^{\circ} \text{C/molal}}$

SIDE NOTE The value of the cryoscopic constant for benzene is actually ${5.12}^{\circ} \text{C/molal}$, so you should check the accuracy of the values given to you.