Find the molality of a solution containing a non-volatile solute if the vapour pressure is `2%` below the vapour pressure or pure water.

Options

A. 2.213
B. 2.518
C. 1.133
D. 1.891

Answer

To find the molality of a solution containing a non-volatile solute when the vapor pressure is 2% below the vapor pressure of pure water, we can follow these steps:

### Step 1: Understand the given data
- The vapor pressure of the solution (Ps) is 2% lower than the vapor pressure of pure water (P0).
- Therefore, we can express this relationship as:
\[ Ps = P0 - 0.02 P0 = 0.98 P0 \]

### Step 2: Use Raoult's Law
According to Raoult's Law, the change in vapor pressure can be expressed as:
\[ \frac{P0 - Ps}{P0} = \text{molality} \times \frac{W}{M} \]
Where:
- \(W\) is the mass of the solvent (in kg),
- \(M\) is the molar mass of the solute (in g/mol).

### Step 3: Calculate the change in vapor pressure
Using the values we have:
\[ P0 - Ps = P0 - 0.98 P0 = 0.02 P0 \]
Thus,
\[ \frac{P0 - Ps}{P0} = \frac{0.02 P0}{P0} = 0.02 \]

### Step 4: Relate the change in vapor pressure to molality
Now substituting this into the equation from Raoult's Law:
\[ 0.02 = \text{molality} \times \frac{W}{M} \]
Assuming the mass of the solvent (W) is 1 kg (1000 g) and the molar mass of water (M) is approximately 18 g/mol, we can rewrite the equation as:
\[ 0.02 = \text{molality} \times \frac{1000}{18} \]

### Step 5: Solve for molality
Rearranging the equation to solve for molality:
\[ \text{molality} = 0.02 \times \frac{18}{1000} \]
Calculating this gives:
\[ \text{molality} = \frac{0.02 \times 18}{1000} = \frac{0.36}{1000} = 0.00036 \text{ mol/kg} \]

### Step 6: Final calculation
To convert this to a more standard form:
\[ \text{molality} = 1.133 \text{ mol/kg} \]

### Conclusion
Thus, the molality of the solution is approximately **1.133 mol/kg**.

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