Question

a certain substance x has a normal freezing point of -7.2 °c and a molal freezing point depression constant ( k_f = 2.73 , ^circ\text{c·kg·mol}^{-1} ). a solution is prepared by dissolving some urea ((\text{(nh}_2\text{)}_2\text{co})) in 350. g of x. this solution freezes at -11.5 °c. calculate the mass of urea that was dissolved. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate freezing point depression

The freezing point depression $\Delta T_f$ is the difference between the freezing point of the pure solvent and the solution. So $\Delta T_f = T_{f, pure} - T_{f, solution}$. Given $T_{f, pure} = -7.2^\circ C$ and $T_{f, solution} = -11.5^\circ C$, then $\Delta T_f = -7.2 - (-11.5) = 4.3^\circ C$.

Step2: Use freezing point depression formula to find molality

The formula for freezing point depression is $\Delta T_f = K_f \cdot m$, where $m$ is molality. Rearranging for $m$, we get $m = \frac{\Delta T_f}{K_f}$. Given $K_f = 2.73^\circ C \cdot kg \cdot mol^{-1}$ and $\Delta T_f = 4.3^\circ C$, so $m = \frac{4.3}{2.73} \approx 1.575 mol/kg$.

Step3: Calculate moles of urea

Molality $m$ is moles of solute per kilogram of solvent. The mass of solvent (substance X) is $350. g = 0.350 kg$. So moles of urea $n = m \cdot mass_{solvent (kg)} = 1.575 mol/kg \cdot 0.350 kg \approx 0.55125 mol$.

Step4: Calculate mass of urea

The molar mass of urea $((NH_2)_2CO)$: $N: 2 \times 14.01 = 28.02$, $H: 4 \times 1.008 = 4.032$, $C: 12.01$, $O: 16.00$. Total molar mass $M = 28.02 + 4.032 + 12.01 + 16.00 = 60.062 g/mol$. Mass of urea $m = n \cdot M = 0.55125 mol \cdot 60.062 g/mol \approx 33.1 g$.

Answer:

33.1 g