The vapours pressure of water at 23°C is 19.8 mm. of Hg. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm of Hg) of the resultant solution?
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 19.602.
To calculate the vapor pressure of the resultant solution, we can use Raoult's law, which states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution.
Given:
Vapor pressure of water (pure solvent) = 19.8 mmHg
Mole fraction of glucose (solute) = 0.1
Mass of water = 178.2 g
First, we need to calculate the mole fraction of water (solvent).
Mole fraction of water ($X_{\text{water}}$) = \(\frac{{\text{moles of water}}}{{\text{total moles}}}\)
Since we are given that 0.1 mole of glucose is dissolved in 178.2 g of water, we can calculate the moles of water:
Moles of water = \(\frac{{\text{mass of water}}}{{\text{molar mass of water}}}\)
The molar mass of water (\(H_2O\)) is approximately 18.02 g/mol.
Moles of water = \(\frac{{178.2 \, \text{g}}}{{18.02 \, \text{g/mol}}} \approx 9.888 \, \text{mol}\)
Total moles = Moles of glucose + Moles of water
Total moles = 0.1 mol + 9.888 mol = 9.988 mol
Now, we can calculate the mole fraction of water:
Mole fraction of water ($X_{\text{water}}$) = \(\frac{{9.888 \, \text{mol}}}{{9.988 \, \text{mol}}} \approx 0.990\)
According to Raoult's law, the vapor pressure of the resultant solution is equal to the mole fraction of water multiplied by the vapor pressure of pure water.
Vapor pressure of the solution = $X_{\text{water}} \times$ Vapor pressure of water
Vapor pressure of the solution = \(0.990 \times 19.8 \, \text{mmHg} \approx 19.602 \, \text{mmHg}\)
Therefore, the vapor pressure of the resultant solution is approximately 19.602 mmHg.
Hence, the correct option is (2) 19.602.