The pressure of the water vapour of a solution containing a nonvolatile solute is 2% below that of the vapour of pure water. Calculate the molality of the solution.

Answer

To solve the problem of calculating the molality of a solution containing a non-volatile solute, we can follow these steps:

### Step 1: Understand the given information
We know that the vapor pressure of the solution (P_s) is 2% lower than that of pure water (P°). The vapor pressure of pure water (P°) is given as 1 atm.

### Step 2: Calculate the vapor pressure of the solution
Since the vapor pressure of the solution is 2% lower than that of pure water, we can calculate it as follows:
\[ P_s = P° - (0.02 \times P°) = P° \times (1 - 0.02) = 1 \, \text{atm} \times 0.98 = 0.98 \, \text{atm} \]

### Step 3: Apply Raoult's Law
According to Raoult's Law, the relationship between the vapor pressures is given by:
\[ \frac{P° - P_s}{P_s} = \text{molality} \times \frac{M_w}{1000} \]
Where:
- \(P°\) = vapor pressure of pure solvent (1 atm)
- \(P_s\) = vapor pressure of the solution (0.98 atm)
- \(M_w\) = molar mass of the solvent (water, which is 18 g/mol)

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ \frac{1 - 0.98}{0.98} = \text{molality} \times \frac{18}{1000} \]
This simplifies to:
\[ \frac{0.02}{0.98} = \text{molality} \times \frac{18}{1000} \]

### Step 5: Calculate the left-hand side
Calculating the left-hand side:
\[ \frac{0.02}{0.98} \approx 0.02041 \]

### Step 6: Rearranging to find molality
Now, we can rearrange the equation to solve for molality (m):
\[ 0.02041 = \text{molality} \times \frac{18}{1000} \]
\[ \text{molality} = \frac{0.02041 \times 1000}{18} \]

### Step 7: Calculate the molality
Calculating the right-hand side:
\[ \text{molality} \approx \frac{20.41}{18} \approx 1.13 \, \text{molal} \]

### Final Answer
The molality of the solution is approximately **1.13 molal**.

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