# Question 5de6f

May 22, 2015

The freezing point of the solution will be equal to ${0.53}^{\circ} \text{C}$.

SIDE NOTE The cryoscopic constant, ${K}_{f}$, for benzene is actually equal to ${5.12}^{\circ} \text{C/molal}$, not ${4.90}^{\circ} \text{C/molal}$, so I solved the problem using the correct value.

If you want, you can redo the calculations with the value you have/were given.

So, you know that you're dealing with a solution containing naphthalene, your solute, and benzene, your solvent. The first thing you need to do is determine the mass of benzene by using its volume and density

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

${m}_{\text{benzene" = 0.877"g"/cancel("mL") * 722cancel("mL") = "633.2 g}}$

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

${K}_{f}$ - the cryoscopic constant - depends on the solvent;
$b$ - 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 - is defined as
${T}_{\text{f"^0 - T_"f sol}}$.

In your case, the van't Hoff factor will be equal to 1 because you'rea dealing with molecular compounds which do not dissociate.

Use naphthalene's molar mass to determine how many moles you're adding to the solution

78.8cancel("g") * "1 mole"/(128.16cancel("g")) = "0.6149 moles"

The solution's molality will thus be

$b = {n}_{\text{naphthalene"/m_"benzene}}$

$b = \text{0.6149 moles"/(633.2 * 10^(-3)"kg") = "0.971 molal}$

Now you have all you need to solve for $\Delta {T}_{\text{f}}$

$\Delta {T}_{\text{F" = 1 * 5.12^@"C"/cancel("molal") * 0.971cancel("molal") = 4.97^@"C}}$

This means that the freezing point of the solution is

$\Delta {T}_{\text{f" = T_"f"^0 - T_"f sol" => T_"f sol" = T_"f"^0 - DeltaT_"f}}$

T_"f sol" = 5.50 - 4.97 = color(green)(0.53^@"C")#