Question

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 .

• A. 4.69 g
• B. 2.59 g
• C. 3.59 g
• D. 3.69 g

Answer 1

The Correct Option is D

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.