# Question da7b4

Jul 26, 2016

${2.42}^{\circ} \text{C}$

#### Explanation:

Your tool of choice here will be the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution

This equation will allow you to find the freezing-point depression, $\Delta {T}_{f}$, for your solution, which in turn will help you find the freezing point of the solution.

Your solute, naphthalene, ${\text{C"_10"H}}_{8}$, is a non-electrolyte, which means that the van't Hoff factor for this solution will be $1$.

The first thing to do here is calculate the molality of the solution by figuring out

• how many moles of solute you have present
• how many kilograms of solvent you have in your solution

The problem provides you the molar masses of the elements that make up naphthalene, so calculate the molar mass of the compound by

M_("M C"_10"H"_8) = 10 xx "12.011 g mol"^(-1) + 8 xx "1.008 g mol"^(-1)

M_("M C"_8"H"_10) = "128.174 g mol"^(-1)

The number of moles of naphthalene present in the sample is equal to

0.134 color(red)(cancel(color(black)("g"))) * ("1 mole C"_10"H"_8)/(128.174 color(red)(cancel(color(black)("g")))) = "0.001045 moles C"_10"H"_8

The mass of cyclohexane, ${\text{C"_6"H}}_{12}$, your solvent, expressed in kilograms, is equal to

5.00 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 5.00 * 10^(-3)"kg"

The molality of the solution will be

$b = {\text{0.001045 moles"/(5.00 * 10^(-3)"kg") = "0.209 mol kg}}^{- 1}$

Next, use your values to calculate $\Delta {T}_{f}$

DeltaT_f = 1 * 20.0^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.209 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))

$\Delta {T}_{f} = {4.18}^{\circ} \text{C}$

Now, a solution's freezing-point depression essentially tells you how the freezing point of the solution, ${T}_{\text{sol}}$, compares to the freezing point of the pure solvent, ${T}_{f}^{\circ}$.

More specifically, the freezing-point depression tells you how many degrees lower the freezing point of the solution will be compared to that of the pure solvent

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{T}_{\text{sol}} = {T}_{f}^{\circ} - \Delta {T}_{f}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

T_"sol" = 6.60^@"C" - 4.18^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(2.42^@"C")color(white)(a/a)|)))#