Sure, let's break down the solution step-by-step.
Given:
- Mass of solute (polymer) = 2.50 g
- Volume of the solution = 150 mL
- Osmotic pressure (Π) = 1.25 × 10^-2 atm
- Temperature = 25°C
We aim to find the molar mass of the polymer.
Step 1: Convert the volume from mL to L
[tex]\[ V = \frac{150 \text{ mL}}{1000} = 0.15 \text{ L} \][/tex]
Step 2: Convert the temperature from Celsius to Kelvin
[tex]\[ T = 25^\circ C + 273.15 = 298.15 \text{ K} \][/tex]
Step 3: Use the osmotic pressure formula to find the molar concentration (C of the solute)
[tex]\[ \Pi = CRT \][/tex]
Where:
- Π = osmotic pressure
- R = ideal gas constant = 0.0821 L·atm/(K·mol)
- T = temperature in Kelvin
Rearrange the formula to solve for [tex]\( C \)[/tex]:
[tex]\[ C = \frac{\Pi}{RT} \][/tex]
Substitute the given values:
[tex]\[ C = \frac{1.25 \times 10^{-2} \text{ atm}}{0.0821 \text{ L·atm/(K·mol)} \times 298.15 \text{ K}} \][/tex]
[tex]\[ C \approx 0.000510660237 \text{ mol/L} \][/tex]
Step 4: Calculate the moles of solute in the solution
[tex]\[ \text{Moles of solute} = C \times V \][/tex]
[tex]\[ \text{Moles of solute} = 0.000510660237 \text{ mol/L} \times 0.15 \text{ L} \][/tex]
[tex]\[ \text{Moles of solute} \approx 7.659903551 \times 10^{-5} \text{ mol} \][/tex]
Step 5: Calculate the molar mass (M) of the polymer
[tex]\[ \text{Molar Mass} = \frac{\text{Mass of solute}}{\text{Moles of solute}} \][/tex]
[tex]\[ \text{Molar Mass} = \frac{2.50 \text{ g}}{7.659903551 \times 10^{-5} \text{ mol}} \][/tex]
[tex]\[ \text{Molar Mass} \approx 32637.487 \text{ g/mol} \][/tex]
Thus, the molar mass of the polymer is approximately 32,637.487 g/mol.
