Question

problem 3: the freezing point of benzene is lowered by 1.2 °c when 1.50 g of an unknown compound is dissolved in 50 g of benzene. (kf = 5.12 °c·kg/mol). determine the molar mass of the compound.

Explanation:

Step1: Recall the freezing - point depression formula

$\Delta T_f = K_f\times m$
where $\Delta T_f$ is the freezing - point depression, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution. First, find the molality $m$.

Step2: Calculate the molality

Given $\Delta T_f = 1.2^{\circ}C$ and $K_f=5.12^{\circ}C\cdot kg/mol$. From $\Delta T_f = K_f\times m$, we can solve for $m$:
$m=\frac{\Delta T_f}{K_f}=\frac{1.2^{\circ}C}{5.12^{\circ}C\cdot kg/mol}= 0.234375\ mol/kg$

Step3: Calculate the number of moles of the solute

The mass of the solvent (benzene) $m_{solvent}=50\ g = 0.05\ kg$. Molality $m=\frac{n}{m_{solvent}}$, where $n$ is the number of moles of the solute. So $n = m\times m_{solvent}=0.234375\ mol/kg\times0.05\ kg = 0.01171875\ mol$

Step4: Calculate the molar mass

The mass of the solute $m_{solute}=1.50\ g$. Molar mass $M=\frac{m_{solute}}{n}$.
$M=\frac{1.50\ g}{0.01171875\ mol}\approx128\ g/mol$

Answer:

$128\ g/mol$