To calculate the freezing point depression when 1.2 grams of an organic compound with a molar mass of 195.8 g/mol is dissolved in 51 grams of benzene (with a freezing point depression constant [tex]\( K_{fp} = 0.83 \; \degree \text{C} \cdot \text{kg/mol} \)[/tex]), follow these steps:
1. Convert the mass of the solvent from grams to kilograms:
- The mass of the solvent (benzene) given is 51 grams.
- Converting this to kilograms:
[tex]\[
\text{mass of solvent} = \frac{51 \; \text{g}}{1000} = 0.051 \; \text{kg}
\][/tex]
2. Calculate the number of moles of the solute:
- The mass of the solute (compound X) is 1.2 grams.
- Molar mass of the solute is 195.8 g/mol.
- Using the formula:
[tex]\[
\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{1.2 \; \text{g}}{195.8 \; \text{g/mol}} \approx 0.0061287 \; \text{mol}
\][/tex]
3. Calculate the molal concentration (molality) of the solution:
- Molality ([tex]\(m\)[/tex]) is defined as the number of moles of solute per kilogram of solvent.
- Using the moles of solute calculated above and the mass of the solvent in kilograms:
[tex]\[
\text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0061287 \; \text{mol}}{0.051 \; \text{kg}} \approx 0.1201706 \; \text{mol/kg}
\][/tex]
4. Calculate the freezing point depression:
- Freezing point depression ([tex]\(\Delta T_f\)[/tex]) is calculated using the formula:
[tex]\[
\Delta T_f = K_{fp} \times \text{molality}
\][/tex]
- Substituting the values:
[tex]\[
\Delta T_f = 0.83 \; \degree \text{C} \cdot \text{kg/mol} \times 0.1201706 \; \text{mol/kg} \approx 0.0997416 \; \degree \text{C}
\][/tex]
Therefore, the freezing point depression of the solution is approximately [tex]\( 0.0997 \; \degree \text{C} \)[/tex].
