Q.Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

The key idea is Raoult's Law, which states that the vapor pressure of a solvent in a solution containing a non-volatile solute is directly proportional to its mole fraction.

1. Calculate the moles of glucose and water:

   $n_{\text{glucose}} = \frac{25 \text{ g}}{180 \text{ g/mol}} = 0.1389 \text{ mol}$

   $n_{\text{water}} = \frac{450 \text{ g}}{18 \text{ g/mol}} = 25 \text{ mol}$

2. Determine the mole fraction of water ($x_{\text{water}}$):

   $x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{glucose}}} = \frac{25 \text{ mol}}{25 \text{ mol} + 0.1389 \text{ mol}} = \frac{25}{25.1389} = 0.99447$ …

When a non-volatile solute, such as glucose, is dissolved in a solvent like water, the vapor pressure of the solvent decreases. This phenomenon is one of the colligative properties, which depend only on the number of solute particles, not their identity. The underlying principle is Raoult's Law.

Concept and Intuition: Why does vapor pressure decrease?

Imagine a pure liquid in a closed container. Molecules at the surface constantly escape into the gas phase (vaporize) and return to the liquid phase (condense). At equilibrium, the rate of vaporization equals the rate of condensation, and the pressure exerted by the vapor is the vapor pressure of the pure liquid.

Now, if we dissolve a non-volatile solute (like glucose) in this liquid, the solute molecules occupy some of the surface area of the liquid. This means fewer solvent molecules are exposed at the surface. With fewer solvent molecules at the surface, the rate at which solvent molecules can escape into the vapor phase decreases. The rate of condensation, however, remains largely unaffected initially. To re-establish equilibrium, the concentration of solvent molecules in the vapor phase must decrease, leading to a lower vapor pressure above the solution compared to the pure solvent.

Raoult's Law quantifies this observation: the partial vapor pressure of each volatile component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. For a solution with a non-volatile solute, only the solvent contributes to the vapor pressure.

Let's apply this to the given problem.

Step-by-step Solution:

1. Identify the given information and what needs to be calculated.

   • Vapor pressure of pure water ($P_A^0$) at 293 K = 17.535 mm Hg.
   • Mass of glucose (solute) = 25 g.
   • Mass of water (solvent) = 450 g.
   • We need to calculate the vapor pressure of water in the solution ($P_A$).

2. Calculate the number of moles of glucose (solute).

   To use Raoult's Law, we need mole fractions. First, we find the moles of glucose.

   The molar mass of glucose ($C_6H_{12}O_6$) is:

   $6 \times 12.01 \text{ g/mol (C)} + 12 \times 1.008 \text{ g/mol (H)} + 6 \times 16.00 \text{ g/mol (O)} = 180.156 \text{ g/mol}$.

   We can round this to $180.16 \text{ g/mol}$ for calculations.

   Moles of glucose ($n_{\text{glucose}}$) = $\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}$

   $n_{\text{glucose}} = \frac{25 \text{ g}}{180.16 \text{ g/mol}} = 0.13876 \text{ mol}$

3. Calculate the number of moles of water (solvent).

   The molar mass of water ($H_2O$) is:

   $2 \times 1.008 \text{ g/mol (H)} + 1 \times 16.00 \text{ g/mol (O)} = 18.016 \text{ g/mol}$.

   We can round this to $18.02 \text{ g/mol}$ for calculations.

   Moles of water ($n_{\text{water}}$) = $\frac{\text{Mass of water}}{\text{Molar mass of water}}$

   $n_{\text{water}} = \frac{450 \text{ g}}{18.02 \text{ g/mol}} = 24.972 \text{ mol}$

4. Calculate the mole fraction of water ($X_{\text{water}}$) in the solution. …

Method: Raoult’s Law for Non-Volatile Solute

Raoult’s Law states that for a solution containing a non-volatile solute (like glucose), the vapour pressure of the solvent above the solution is directly proportional to its mole fraction in the solution.

Where:

• $P_{\text{solution}}$ = vapour pressure of the solution
• $X_{\text{solvent}}$ = mole fraction of the solvent (water)
• $P_{\text{solvent}}^0$ = vapour pressure of pure solvent at the same temperature

Steps

Step 1: Identify given data

• $P_{\text{water}}^0 = 17.535 \text{ mm Hg}$ at 293 K
• Mass of glucose = 25 g
• Mass of water = 450 g
• Molar mass of glucose ($C_6H_{12}O_6$) = $6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \text{ g/mol}$
• Molar mass of water ($H_2O$) = $2 \times 1 + 16 = 18 \text{ g/mol}$

Step 2: Calculate moles of each component

Step 3: Calculate mole fraction of water

$X_{\text{water}} = \frac{25}{25 + 0.1389} = \frac{25}{25.1389} \approx 0.9945$ …

Here are the common mistakes students make on this exact type of problem (vapour pressure of water after dissolving a non-volatile solute), and how to avoid each.

1. Forgetting to Convert Mass to Moles First

The Mistake: Using the given masses (25 g glucose, 450 g water) directly instead of converting to moles.

How to avoid: Mass -> moles -> mole fraction -> Raoult's law.

• Molar mass of glucose ($C_6H_{12}O_6$) = $6(12)+12(1)+6(16) = 180$ g/mol -> moles $= \dfrac{25}{180} \approx 0.139$ mol
• Molar mass of water = 18 g/mol -> moles $= \dfrac{450}{18} = 25$ mol

2. Using the Mole Fraction of Glucose Instead of Water

The Mistake: Writing $P = x_{\text{glucose}} \cdot P^\circ_{\text{water}}$.

Why it's wrong: Glucose is non-volatile, so it contributes nothing to the vapour phase -- the solution's vapour pressure comes entirely from the solvent (water), scaled by the solvent's own mole fraction.

How to avoid: Always use $P_{\text{solution}} = x_{\text{water}} \cdot P^\circ_{\text{water}}$, not $x_{\text{solute}}$.

3. Forgetting the Solute in the Denominator of the Mole Fraction

The Mistake: Computing $x_{\text{water}} = 1$ by ignoring the small amount of glucose.

How to avoid: $x_{\text{water}} = \dfrac{n_{\text{water}}}{n_{\text{water}}+n_{\text{glucose}}}$ -- even though glucose's contribution is small, it must be included; otherwise you'd predict zero vapour pressure lowering.

4. Rounding Too Early …