Question

a certain substance x has a normal boiling point of 133.5 °c and a molal boiling point elevation constant ( k_b = 1.37 , ^circ\text{c·kg·mol}^{-1} ). a solution is prepared by dissolving some urea ((\text{nh}_2)_2\text{co}) in 900. g of x. this solution boils at 134.8 °c. calculate the mass of urea that was dissolved. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate boiling point elevation

The boiling point elevation $\Delta T_b$ is the difference between the boiling point of the solution and the boiling point of the pure solvent. So $\Delta T_b = T_{solution} - T_{pure}$.
Given $T_{solution} = 134.8^{\circ}C$ and $T_{pure} = 133.5^{\circ}C$, then $\Delta T_b = 134.8 - 133.5 = 1.3^{\circ}C$.

Step2: Use boiling point elevation formula to find molality

The formula for boiling point elevation is $\Delta T_b = K_b \times m$, where $m$ is the molality. We can solve for $m$: $m = \frac{\Delta T_b}{K_b}$.
Given $K_b = 1.37^{\circ}C \cdot kg \cdot mol^{-1}$ and $\Delta T_b = 1.3^{\circ}C$, then $m = \frac{1.3}{1.37} \approx 0.9489 mol/kg$.

Step3: Calculate moles of urea

Molality $m$ is defined as moles of solute per kilogram of solvent. The mass of solvent (substance X) is $900. g = 0.900 kg$. Let $n$ be the moles of urea. Then $m = \frac{n}{mass_{solvent}(kg)}$, so $n = m \times mass_{solvent}(kg)$.
Substituting the values: $n = 0.9489 mol/kg \times 0.900 kg \approx 0.8540 mol$.

Step4: Calculate molar mass of urea

The formula for urea is $(NH_2)_2CO$. The molar mass $M$ is calculated as:
$M = 2 \times (14.01 + 2 \times 1.008) + 12.01 + 16.00$
$= 2 \times (14.01 + 2.016) + 12.01 + 16.00$
$= 2 \times 16.026 + 12.01 + 16.00$
$= 32.052 + 12.01 + 16.00 = 60.062 g/mol$.

Step5: Calculate mass of urea

Mass $m_{urea} = n \times M$.
Substituting $n = 0.8540 mol$ and $M = 60.062 g/mol$: $m_{urea} = 0.8540 mol \times 60.062 g/mol \approx 51.3 g$.

Answer:

$\boxed{51.3}$ (in grams, with appropriate significant figures)