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 molal 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 1: Understand the Concept of Lowering of Vapor Pressure
The lowering of vapor pressure can be calculated using the formula:
\[ \text{Relative Lowering of Vapor Pressure} = \frac{P_0 - P}{P_0} = \text{Mole Fraction of Solute} \]
Where \( P_0 \) is the vapor pressure of the pure solvent and \( P \) is the vapor pressure of the solution.

### Step 2: Define the Variables
- Given molality (m) = 1.00 molal
- Molar mass of the solvent (M) = 40 g/mol
- At normal boiling point, the vapor pressure of the solvent \( P_0 \) = 760 mmHg (which is equivalent to 760 Torr).

### Step 3: Calculate the Number of Moles of Solvent
Since molality is defined as the number of moles of solute per kilogram of solvent, we can express the number of moles of solvent in terms of its mass:
\[ \text{Number of moles of solvent} (N) = \frac{\text{mass of solvent (in grams)}}{\text{molar mass of solvent (g/mol)}} \]
Assuming we have 1 kg (1000 g) of solvent:
\[ N = \frac{1000 \text{ g}}{40 \text{ g/mol}} = 25 \text{ moles} \]

### Step 4: Calculate the Number of Moles of Solute
Since the solution is 1.00 molal, it means there is 1 mole of solute in 1 kg of solvent.

### Step 5: Calculate the Mole Fraction of Solute
The mole fraction of solute (\( X_{solute} \)) can be calculated as:
\[ X_{solute} = \frac{\text{Number of moles of solute}}{\text{Number of moles of solute} + \text{Number of moles of solvent}} = \frac{1}{1 + 25} = \frac{1}{26} \]

### Step 6: Calculate the Lowering of Vapor Pressure
Using the formula for lowering of vapor pressure:
\[ P_0 - P = P_0 \times X_{solute} \]
Substituting the values:
\[ P_0 - P = 760 \text{ Torr} \times \frac{1}{26} \approx 29.23 \text{ Torr} \]

### Step 7: Final Answer
Thus, the lowering of vapor pressure is approximately:
\[ \text{Lowering of Vapor Pressure} \approx 29.23 \text{ Torr} \]