The vapour pressure of pure water (molar mass 18.0 g/mole) at 50 oC is 92.5 mm Hg. A solution containing the non-electrolyte sucrose (molar mass 342 g/mole) has a vapour pressure of 90.8 mm Hg at 50 °C. The vapour pressure of pure water (molar a. What is the boiling point elevation (Ts) of this solution (Ks(water) = 0.520 °C/m)? b. If it has a density of 1.08 g/mL, what is the osmotic pressure of this solution at 50 °C?
The Correct Answer and Explanation is :
To solve this, let’s address both parts step by step:
Part a: Boiling Point Elevation (( \Delta T_b ))
- Determine the mole fraction of sucrose:
The lowering of vapor pressure is proportional to the mole fraction of the solute:
[
\Delta P = P_{\text{pure}} – P_{\text{solution}} = 92.5 – 90.8 = 1.7 \, \text{mm Hg}
]
The mole fraction of water ((X_{\text{water}})) is:
[
X_{\text{water}} = \frac{P_{\text{solution}}}{P_{\text{pure}}} = \frac{90.8}{92.5} = 0.982
]
The mole fraction of sucrose ((X_{\text{sucrose}})) is:
[
X_{\text{sucrose}} = 1 – X_{\text{water}} = 1 – 0.982 = 0.018
] - Calculate the molality ((m)):
Assume 1 kg of water. Using ( X_{\text{sucrose}} ):
[
X_{\text{sucrose}} = \frac{\text{moles of sucrose}}{\text{moles of sucrose} + \text{moles of water}}
]
Water moles (( n_{\text{water}} )):
[
n_{\text{water}} = \frac{1000}{18} = 55.56 \, \text{mol}
]
Let ( n_{\text{sucrose}} ) represent moles of sucrose:
[
X_{\text{sucrose}} \approx \frac{n_{\text{sucrose}}}{55.56} \implies n_{\text{sucrose}} = 0.018 \times 55.56 = 1.0 \, \text{mol}
]
Since (1.0 \, \text{mol sucrose}) is in (1 \, \text{kg water}), the molality is:
[
m = 1.0 \, \text{mol/kg}
] - Calculate ( \Delta T_b ):
Using the boiling point elevation formula:
[
\Delta T_b = K_b \cdot m = 0.520 \, \text{°C/m} \times 1.0 \, \text{mol/kg} = 0.520 \, \text{°C}
]
Part b: Osmotic Pressure (( \Pi ))
- Determine molarity (( C )):
Assume (1 \, \text{L solution}). The density is (1.08 \, \text{g/mL}), so mass of solution is:
[
\text{Mass of solution} = 1.08 \, \text{g/mL} \times 1000 \, \text{mL} = 1080 \, \text{g}
]
Mass of water (( \text{mass}{\text{water}} )): [ \text{mass}{\text{water}} = 1080 \, \text{g} – (1.0 \, \text{mol} \times 342 \, \text{g/mol}) = 738 \, \text{g}
]
Volume of water is approximately ( 738 \, \text{mL} ), so the solution volume is close to (1 \, \text{L}). Therefore:
[
C \approx 1.0 \, \text{mol/L}
] - Calculate osmotic pressure:
Using the formula ( \Pi = CRT ):
[
R = 0.0821 \, \text{L·atm/(mol·K)}, \quad T = 50 + 273 = 323 \, \text{K}
]
[
\Pi = (1.0 \, \text{mol/L}) \cdot (0.0821) \cdot (323) = 26.5 \, \text{atm}
]
Final Results:
a. Boiling Point Elevation (( \Delta T_b )): ( 0.520 \, \text{°C} )
b. Osmotic Pressure (( \Pi )): ( 26.5 \, \text{atm} )
Explanation (300 words):
The boiling point elevation and osmotic pressure are colligative properties, meaning they depend on the number of solute particles rather than their identity. In part (a), the vapor pressure lowering revealed the mole fraction of sucrose, allowing us to calculate the molality of the solution. Given the boiling point elevation constant (( K_b )) for water, the increase in boiling point was determined as ( 0.520 \, \text{°C} ).
In part (b), osmotic pressure was calculated based on the molarity of sucrose. The solution’s density helped estimate its volume, confirming the molarity as ( 1.0 \, \text{mol/L} ). Using the ideal gas constant and temperature in Kelvin, the osmotic pressure was found to be ( 26.5 \, \text{atm} ). These results illustrate how colligative properties are vital for determining solute behavior in solutions.