`12g` of a nonvolatile solute dissolved in `108g` of water produces the relative lowering of vapour pressure of `0.1`. The molecular mass of the solute is

Options

A. 60
B. 80
C. 40
D. 20

Answer

To find the molecular mass of the non-volatile solute, we can use the concept of relative lowering of vapor pressure. Here’s the step-by-step solution:

### Step 1: Understand the formula for relative lowering of vapor pressure
The relative lowering of vapor pressure is given by the formula:
\[ \frac{P_0 - P_s}{P_0} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
Where:
- \(P_0\) = vapor pressure of pure solvent
- \(P_s\) = vapor pressure of the solution
- \(n_{solute}\) = number of moles of solute
- \(n_{solvent}\) = number of moles of solvent

### Step 2: Simplify for dilute solutions
For dilute solutions, the number of moles of solute is much smaller than that of the solvent, so we can approximate:
\[ \frac{P_0 - P_s}{P_0} \approx \frac{n_{solute}}{n_{solvent}} \]

### Step 3: Substitute the known values
We know from the problem statement that the relative lowering of vapor pressure is given as \(0.1\):
\[ 0.1 = \frac{n_{solute}}{n_{solvent}} \]

### Step 4: Calculate moles of solvent
The mass of the solvent (water) is given as \(108g\). The molecular mass of water is \(18 g/mol\). Therefore, the number of moles of solvent can be calculated as:
\[ n_{solvent} = \frac{mass_{solvent}}{molar mass_{solvent}} = \frac{108g}{18g/mol} = 6 mol \]

### Step 5: Calculate moles of solute
Using the relationship from step 3:
\[ 0.1 = \frac{n_{solute}}{6} \]
Rearranging gives:
\[ n_{solute} = 0.1 \times 6 = 0.6 mol \]

### Step 6: Relate moles of solute to mass and molecular mass
We know that:
\[ n_{solute} = \frac{mass_{solute}}{molar mass_{solute}} \]
Given that the mass of the solute is \(12g\), we can write:
\[ 0.6 = \frac{12g}{molar mass_{solute}} \]

### Step 7: Solve for the molecular mass of the solute
Rearranging the equation gives:
\[ molar mass_{solute} = \frac{12g}{0.6} = 20 g/mol \]

### Final Answer
The molecular mass of the solute is \(20 g/mol\).