Questions: When 183. mg of a certain molecular compound X are dissolved in 50.0 g of benzene (C6H6), the freezing point of the solution is measured to be 5.2°C. Calculate the molar mass of X.

When 183. mg of a certain molecular compound X are dissolved in 50.0 g of benzene (C6H6), the freezing point of the solution is measured to be 5.2°C. Calculate the molar mass of X.

Transcript text: When 183. mg of a certain molecular compound $X$ are dissolved in 50.0 g of benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$, the freezing point of the solution is measured to be $5.2^{\circ} \mathrm{C}$. Calculate the molar mass of $X$.

Solution

Solution Steps

Step 1: Determine the Freezing Point Depression

The freezing point of pure benzene is $5.5^\circ \text{C}$. The freezing point of the solution is given as $5.2^\circ \text{C}$. The freezing point depression ($\Delta T_f$) is calculated as:

\[ \Delta T_f = 5.5^\circ \text{C} - 5.2^\circ \text{C} = 0.3^\circ \text{C} \]

Step 2: Use the Freezing Point Depression Formula

The formula for freezing point depression is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

where:

$i$ is the van't Hoff factor (which is 1 for non-electrolytes),
$K_f$ is the cryoscopic constant for benzene, $5.12 \, \text{°C kg/mol}$,
$m$ is the molality of the solution.

Rearranging the formula to solve for molality ($m$):

\[ m = \frac{\Delta T_f}{i \cdot K_f} = \frac{0.3^\circ \text{C}}{1 \cdot 5.12 \, \text{°C kg/mol}} = 0.0586 \, \text{mol/kg} \]

Step 3: Calculate the Moles of Solute

The molality ($m$) is defined as the moles of solute per kilogram of solvent. Given that the mass of benzene is 50.0 g (or 0.0500 kg), we can find the moles of solute ($n$):

\[ n = m \cdot \text{mass of solvent (kg)} = 0.0586 \, \text{mol/kg} \times 0.0500 \, \text{kg} = 0.00293 \, \text{mol} \]

Step 4: Calculate the Molar Mass of the Solute

The molar mass ($M$) of the solute is calculated using the mass of the solute and the moles of solute:

\[ M = \frac{\text{mass of solute (g)}}{\text{moles of solute (mol)}} = \frac{183 \, \text{mg}}{0.00293 \, \text{mol}} = \frac{0.183 \, \text{g}}{0.00293 \, \text{mol}} = 62.5 \, \text{g/mol} \]