To determine the freezing point of an aqueous solution of a nonvolatile non-electrolyte with the given properties, follow these steps:
1. Determine the Molarity (M) of the Solution:
The osmotic pressure (π) is given by the formula:
[tex]\[
\pi = MRT
\][/tex]
where:
- [tex]\( \pi \)[/tex] is the osmotic pressure (2.0 atm),
- [tex]\( M \)[/tex] is the molarity of the solution (in mol/L),
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L atm K⁻¹ mol⁻¹),
- [tex]\( T \)[/tex] is the temperature in Kelvin (300 K).
Rearrange the formula to solve for molarity [tex]\( M \)[/tex]:
[tex]\[
M = \frac{\pi}{RT}
\][/tex]
Substitute the known values:
[tex]\[
M = \frac{2.0}{0.0821 \times 300} ≈ 0.0812 \, \text{mol/L}
\][/tex]
2. Assume Molality (m) is approximately equal to Molarity (M):
For dilute aqueous solutions, the molarity [tex]\( M \)[/tex] and molality [tex]\( m \)[/tex] can be considered approximately equal because the density of water is close to 1 kg/L. Hence, we assume:
[tex]\[
m \approx M ≈ 0.0812 \, \text{mol/kg}
\][/tex]
3. Calculate the Freezing Point Depression (ΔTf):
The freezing point depression is given by:
[tex]\[
\Delta T_f = K_f \times m
\][/tex]
where [tex]\( K_f \)[/tex] is the freezing point depression constant (1.86 K kg mol⁻¹).
Substitute the known values:
[tex]\[
\Delta T_f = 1.86 \times 0.0812 ≈ 0.151 \, \text{K}
\][/tex]
4. Determine the Freezing Point of the Solution:
The freezing point of pure water is 0°C, which is 273 K. The freezing point of the solution is given by:
[tex]\[
T_{f,\text{solution}} = 273 \, \text{K} - \Delta T_f
\][/tex]
Substitute the calculated value of [tex]\( \Delta T_f \)[/tex]:
[tex]\[
T_{f,\text{solution}} = 273 - 0.151 ≈ 272.849 \, \text{K}
\][/tex]
Thus, the freezing point of the aqueous solution is approximately 272.849 K.
