High School

Perylene [tex] \left( C_{20}H_{12} \right) [/tex] is a constituent of coal tar. How many grams of perylene must be dissolved in 66.9 g of chloroform in order to lower the freezing point by [tex] 2.75^\circ C [/tex]? The [tex] K_f [/tex] for chloroform is [tex] 4.68^\circ C/m [/tex].

Answer :

To find out how many grams of perylene must be dissolved in 66.9 g of chloroform to lower the freezing point by [tex]\( 2.75^\circ \text{C} \)[/tex], we can follow these steps:

1. Identify the Given Values:
- Freezing point depression ([tex]\(\Delta T_f\)[/tex]) = [tex]\( 2.75^\circ \text{C} \)[/tex]
- Cryoscopic constant for chloroform ([tex]\(K_f\)[/tex]) = [tex]\( 4.68^\circ \text{C/m}\)[/tex]
- Mass of chloroform = [tex]\( 66.9 \text{ g} \)[/tex]

2. Calculate the Molality ([tex]\(m\)[/tex]) of the Solution:
Molality is calculated using the formula:
[tex]\[
m = \frac{\Delta T_f}{K_f}
\][/tex]
Substituting the values, we get:
[tex]\[
m = \frac{2.75}{4.68} \approx 0.5876 \text{ mol/kg}
\][/tex]

3. Determine the Molar Mass of Perylene ([tex]\(C_{20}H_{12}\)[/tex]):
- Carbon ([tex]\(C\)[/tex]) has a molar mass of [tex]\( 12.01 \text{ g/mol} \)[/tex]
- Hydrogen ([tex]\(H\)[/tex]) has a molar mass of [tex]\( 1.008 \text{ g/mol} \)[/tex]

The molar mass of perylene is calculated as:
[tex]\[
\text{Molar Mass of Perylene} = (20 \times 12.01) + (12 \times 1.008) = 240.2 + 12.096 = 252.296 \text{ g/mol}
\][/tex]

4. Calculate the Moles of Perylene Required:
Molality is defined as the moles of solute per kilogram of solvent. We can find the moles of perylene using:
[tex]\[
\text{Moles of Perylene} = m \times \frac{\text{Mass of Chloroform in kg}}{\text{kg of solvent}}
\][/tex]
Converting the mass of chloroform to kilograms:
[tex]\[
66.9 \text{ g} = 0.0669 \text{ kg}
\][/tex]
Therefore:
[tex]\[
\text{Moles of Perylene} = 0.5876 \text{ mol/kg} \times 0.0669 \text{ kg} \approx 0.03931 \text{ mol}
\][/tex]

5. Convert Moles of Perylene to Grams:
The grams of perylene can be determined by:
[tex]\[
\text{Grams of Perylene} = \text{Moles of Perylene} \times \text{Molar Mass of Perylene}
\][/tex]
[tex]\[
\text{Grams of Perylene} = 0.03931 \text{ mol} \times 252.296 \text{ g/mol} \approx 9.918 \text{ g}
\][/tex]

Thus, approximately 9.918 grams of perylene must be dissolved in 66.9 grams of chloroform to achieve the desired freezing point depression.