# Question 71a55

May 12, 2015

Your compound will have a molar mass of 200 g/mol.

To solve this problem, you're going to need the ebullioscopic constant, ${K}_{b}$, and the boling temperature of pure carbon tetrachloride, $C C {l}_{4}$.

${K}_{b} = {5.03}^{\circ} {\text{C kg mol}}^{- 1}$

${T}_{\text{b"^0 = 76.5^@"C}}$

Notice that there's a difference between the boiling point of the pure solvent and the boling point of the solution, which implies that you can use the boiling-point elevation formula

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

$\Delta {T}_{\text{b}}$ - the poiling point elevation;
$i$ - the van't Hoff factor;
${K}_{b}$ - the ebullioscopic constant;
$b$ - the molality of the solution.

Molality is defined as moles of solute, in your case the unknown compound, per kilograms of solvent, in your case carbon tetrachloride.

In general, the van't Hoff factor for substances soluble in non-polar solvents is equal to 1.

Use the above equation to solve for the molality of the solution

$\Delta {T}_{b} = i \cdot {K}_{b} \cdot b \implies b = \frac{\Delta {T}_{\text{b}}}{{K}_{b}}$

b = (T_"b sol" - T_"b"^0)/K_b = ((81.5 - 76.5)""^@cancel("C"))/(5.03""^@cancel("C")"kg mol"^(-1)) = "0.994 mol/kg"

Use the equation for molality to determine how many moles of the unknown compound you have present

b = n/m_"solvent" => n = b * underbrace(m_"solvent")_(color(blue)("in kilograms!"))#

$n = 0.994 \text{mol"/cancel("kg") * 25 * 10^(-3)cancel("kg") = "0.02485 moles}$

Now use the compound's mass to solve for its molar mass

${M}_{M} = \frac{m}{n} = \text{5 g"/"0.02485 moles" = "201.2 g/mol}$

Rounded to one sig fig, the number of sig figs you gave for the mass of the compound, the answer will be

${M}_{M} = \textcolor{g r e e n}{\text{200 g/mol}}$