What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol–1 .
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol–1 .
- 4.69 g
- 2.59 g
- 3.59 g
- 3.69 g
The Correct Option is D
Solution and Explanation
Step 1: Formula for Relative Lowering of Vapor Pressure
For a dilute solution, the relative lowering of vapor pressure is given by:
\[ \frac{P_0 - P_s}{P_0} = \frac{n}{N}, \]
where:
- \( P_0 \): Vapor pressure of the pure solvent
- \( P_s \): Vapor pressure of the solution
- \( n \): Number of moles of solute
- \( N \): Number of moles of solvent
Step 2: Substitute the Given Values
Given:
- \( \frac{P_0 - P_s}{P_0} = \frac{0.2}{54.2} \)
- \( N = 100 \, \text{moles} \)
Substitute into the equation:
\[ \frac{0.2}{54.2} = \frac{n}{100}. \]
Step 3: Solve for \( n \)
Rearranging and solving for \( n \):
\[ n = \frac{100 \times 0.2}{54.2} = \frac{20}{54.2} \approx 0.369 \, \text{moles}. \]
Step 4: Calculate the Mass of the Solute (\( w \))
The mass of the solute is given by:
\[ w = n \times M, \]
where \( M = 180 \, \text{g/mol} \) is the molar mass of the solute. Substituting the values:
\[ w = 0.369 \times 180 \approx 3.69 \, \text{g}. \]
Final Answer:
The mass of the solute is \( w = 3.69 \, \text{g} \).
In vapor pressure calculations, ensure the solution is dilute and units for molar mass and pressure are consistent.
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