Question

calculate the boiling point ($t_{b}$), in degrees celsius, for a solution that contains 122.2 g of $alcl_{3}$ and 457 g of water. $k_{b}$ for water is 0.512 °c/m.

Explanation:

Step1: Calculate moles of AlCl₃

Molar mass of AlCl₃ = 26.98 + 3×35.45 = 133.33 g/mol
$n_{AlCl_3}=\frac{122.2\ g}{133.33\ g/mol}\approx0.917\ mol$

Step2: Calculate molality (m) of the solution

Mass of water in kg = 457 g = 0.457 kg
$m=\frac{n_{AlCl_3}}{mass\ of\ water\ (kg)}=\frac{0.917\ mol}{0.457\ kg}\approx2.01\ m$

Step3: Determine the van't Hoff factor (i) for AlCl₃

AlCl₃ dissociates into Al³⁺ and 3 Cl⁻, so i = 4

Step4: Calculate the boiling - point elevation (ΔT_b)

$\Delta T_b = i\times K_b\times m$
$\Delta T_b=4\times0.512^{\circ}C/m\times2.01\ m\approx4.11^{\circ}C$

Step5: Calculate the boiling point of the solution

Boiling point of pure water, $T_{b0}=100^{\circ}C$
$T_b=T_{b0}+\Delta T_b = 100^{\circ}C + 4.11^{\circ}C=104.11^{\circ}C$

Answer:

$104.11^{\circ}C$