Question

Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at 100°C is:
(Vapour pressure of pure water at 100°C = 760 torr)

• A. 76.0 torr
• B. 752.0 torr
• C. 7.6 torr
• D. 3207.6 torr

Hint:

The calculation involves the mole fractions and vapor pressure relationship for the solute and solvent.

Answer 1

The Correct Option is B

To determine the vapour pressure of the solution containing glucose and water at 100°C, we can use Raoult's Law. According to Raoult's Law, the vapour pressure of a solution is the product of the mole fraction of the solvent and the vapour pressure of the pure solvent. Mathematically, it is given by:

\(P_{\text{solution}} = \chi_{\text{solvent}} \cdot P_{\text{pure solvent}}\) 

Where:

• \(P_{\text{solution}}\) is the vapour pressure of the solution.
• \(\chi_{\text{solvent}}\) is the mole fraction of the solvent (water).
• \(P_{\text{pure solvent}}\) is the vapour pressure of pure water at 100°C, given as 760 torr.

First, we need to calculate the moles of glucose and water:

• Molar mass of glucose (C₆H₁₂O₆): 180 g/mol
• Moles of glucose: \(\frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ mol}\)
• Molar mass of water (H₂O): 18 g/mol
• Moles of water: \(\frac{178.2 \text{ g}}{18 \text{ g/mol}} = 9.9 \text{ mol}\)

Now, calculate the mole fraction of water:

\(\chi_{\text{water}} = \frac{\text{Moles of water}}{\text{Moles of glucose} + \text{Moles of water}} = \frac{9.9}{0.1 + 9.9} = \frac{9.9}{10} = 0.99\)

Using Raoult's Law, calculate the vapour pressure of the solution:

\(P_{\text{solution}} = \chi_{\text{water}} \cdot P_{\text{pure water}} = 0.99 \times 760 \text{ torr} = 752.4 \text{ torr}\)

Considering significant figures and rounding, the vapour pressure of the solution is approximately 752.0 torr.

Hence, the correct answer is: 752.0 torr.

Answer 2

\( n_{H_2O} = \frac{178.2}{18} = 9.9; \, n_{C_6H_{12}O_6} = \frac{18}{180} = 0.1 \)

\( \chi_{C_6H_{12}O_6} = \frac{0.1}{10} = 0.01 \)

\( \frac{P_0 - P_s}{P_0} = \chi_{C_6H_{12}O_6} \)

\( \frac{760 - P_s}{760} = 0.01 \)

\( 760 - P_s = 7.6 \)

\( P_s = 752.4 \, \text{torr} \)