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

Answer

To solve the problem of determining 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 is 2% lower than that of pure water. This means that if the vapor pressure of pure water is \( P^0 \), the vapor pressure of the solution \( P \) can be expressed as:
\[ P = P^0 - 0.02 P^0 = 0.98 P^0 \]

### Step 2: Use Raoult's Law
According to Raoult's Law, the vapor pressure of a solution is related to the mole fraction of the solvent:
\[ P = X_{solvent} \cdot P^0 \]
Where \( X_{solvent} \) is the mole fraction of the solvent.

### Step 3: Relate the Mole Fraction to the Decrease in Vapor Pressure
From the previous step, we can equate the two expressions for \( P \):
\[ 0.98 P^0 = X_{solvent} \cdot P^0 \]
Dividing both sides by \( P^0 \) (assuming \( P^0 \neq 0 \)):
\[ 0.98 = X_{solvent} \]

### Step 4: Calculate the Mole Fraction of the Solute
Since the sum of the mole fractions of the solute and solvent must equal 1:
\[ X_{solute} = 1 - X_{solvent} = 1 - 0.98 = 0.02 \]

### Step 5: Relate Mole Fractions to Molality
Let’s denote the number of moles of solute as \( n_{solute} \) and the number of moles of solvent as \( n_{solvent} \). The mole fraction of the solute can be expressed as:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
Substituting \( X_{solute} = 0.02 \):
\[ 0.02 = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
This implies:
\[ 0.02(n_{solute} + n_{solvent}) = n_{solute} \]
Rearranging gives:
\[ 0.02 n_{solvent} = 0.98 n_{solute} \]
Thus:
\[ \frac{n_{solute}}{n_{solvent}} = \frac{0.02}{0.98} \]

### Step 6: Calculate Molality
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent. If we assume 1 kg (1000 g) of solvent, then:
\[ n_{solvent} = \frac{1000 \, g}{18 \, g/mol} \approx 55.56 \, mol \]
Now substituting back to find \( n_{solute} \):
\[ n_{solute} = \frac{0.02}{0.98} \times n_{solvent} = \frac{0.02}{0.98} \times 55.56 \approx 1.13 \, mol \]
Thus, the molality \( m \) is:
\[ m = \frac{n_{solute}}{mass \, of \, solvent \, in \, kg} = \frac{1.13 \, mol}{1 \, kg} = 1.13 \, mol/kg \]

### Final Answer
The molality of the solution is approximately \( 1.13 \, mol/kg \).

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