The vapour pressure of water at `20^(@)C is 17.54mm`. When 20g of non - ionic substance is dissolved in 100g of water, the vapour pressure is lowered by `0.30mm`. What is the molecular mass of the substance ?

Options

A. 200.8 g/mol
B. 206.88 g/mol
C. 210.5 g/mol
D. 215.2 g/mol

Answer

To solve the problem of finding the molecular mass of the non-ionic substance dissolved in water, we can follow these steps:

### Step 1: Identify the given data
- Vapour pressure of pure water (P₀) = 17.54 mm
- Vapour pressure of the solution (P) = P₀ - 0.30 mm = 17.54 mm - 0.30 mm = 17.24 mm
- Mass of the non-ionic substance (solute) = 20 g
- Mass of water (solvent) = 100 g

### Step 2: Calculate the mole fraction of the solvent
First, we need to calculate the number of moles of water (n₁):
- Molar mass of water (H₂O) = 18 g/mol
- Number of moles of water (n₁) = mass of water / molar mass of water
\[ n₁ = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.56 \text{ moles} \]

### Step 3: Use Raoult's Law
According to Raoult's Law, the lowering of vapour pressure can be expressed as:
\[ \frac{P₀ - P}{P₀} = \frac{n₂}{n₁} \]
where n₂ is the number of moles of solute.

### Step 4: Calculate the change in vapour pressure
Substituting the values we have:
\[ \frac{17.54 \text{ mm} - 17.24 \text{ mm}}{17.54 \text{ mm}} = \frac{n₂}{5.56} \]
Calculating the left side:
\[ \frac{0.30 \text{ mm}}{17.54 \text{ mm}} \approx 0.0171 \]

### Step 5: Calculate n₂
Now we can set up the equation:
\[ 0.0171 = \frac{n₂}{5.56} \]
Solving for n₂:
\[ n₂ = 0.0171 \times 5.56 \approx 0.095 \]

### Step 6: Calculate the molecular mass (M)
We know that:
\[ n₂ = \frac{\text{mass of solute}}{\text{molecular mass (M)}} \]
Rearranging gives:
\[ M = \frac{\text{mass of solute}}{n₂} \]
Substituting the known values:
\[ M = \frac{20 \text{ g}}{0.095} \approx 210.53 \text{ g/mol} \]

### Final Answer
The molecular mass of the non-ionic substance is approximately **210.6 g/mol**.

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