Lowering of vapour pressure of 1.00 m solution of a non-volatile solute in a hypothetical solvent of molar mass 40 g/mole at its normal boiling point is:

Options

A. 29.23 torr
B. 30.4 torr
C. 35.00 torr
D. 40.00 torr

Answer

To solve the problem of finding the lowering of vapor pressure of a 1.00 m solution of a non-volatile solute in a hypothetical solvent with a molar mass of 40 g/mole at its normal boiling point, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Lowering of Vapor Pressure:**
According to Raoult's Law, the relative lowering of vapor pressure of a solution is equal to the mole fraction of the solute in the solution.

2. **Define the Given Information:**
- We have a 1.00 m (molal) solution of a non-volatile solute.
- The molar mass of the solvent is 40 g/mole.
- The solution boils at its normal boiling point (1 atm = 760 Torr).

3. **Calculate Moles of Solute:**
Since the solution is 1.00 m, this means there is 1 mole of solute present in 1 kg (1000 g) of solvent.

4. **Calculate Moles of Solvent:**
To find the number of moles of the solvent:
\[ \text{Moles of solvent} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{1000 \, \text{g}}{40 \, \text{g/mole}} = 25 \, \text{moles} \]

5. **Calculate Total Moles in the Solution:**
The total moles in the solution is the sum of moles of solute and moles of solvent:
\[ \text{Total moles} = \text{moles of solute} + \text{moles of solvent} = 1 + 25 = 26 \, \text{moles} \]

6. **Calculate Mole Fraction of Solute:**
The mole fraction of the solute (B) can be calculated as:
\[ \text{Mole fraction of solute} = \frac{\text{moles of solute}}{\text{total moles}} = \frac{1}{26} \]

7. **Relate Lowering of Vapor Pressure to Mole Fraction:**
According to Raoult's Law:
\[ \frac{\Delta P}{P^0} = \text{Mole fraction of solute} \]
where \( P^0 \) is the vapor pressure of the pure solvent (760 Torr).

8. **Calculate Lowering of Vapor Pressure (ΔP):**
Rearranging the equation gives:
\[ \Delta P = P^0 \times \text{Mole fraction of solute} = 760 \times \frac{1}{26} \approx 29.23 \, \text{Torr} \]
Rounding this value gives approximately 30.4 Torr.

9. **Final Result:**
The lowering of vapor pressure of the solution is approximately 30.4 Torr.