The vapour pressure of pure liquid A is `10` torr and at the same temperature when `1` g solid B is dissolved in `20`g of A , its vapour pressure is reduced to`9.0`torr . If the molecular mass of A is `200`amu , then the molecular mass of B is

Options

A. `100`amu
B. `90`amu
C. `75`amu
D. `120`amu

Explanation

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

### Step 1: Write down the known values
- Vapor pressure of pure liquid A (P₀) = 10 torr
- Vapor pressure of the solution (P) = 9 torr
- Mass of A (m₁) = 20 g
- Mass of B (m₂) = 1 g
- Molecular mass of A (M₁) = 200 amu

### Step 2: Calculate the lowering of vapor pressure
Using the formula for relative lowering of vapor pressure:
\[ \frac{P₀ - P}{P₀} = \frac{n₂}{n₁ + n₂} \]
Where:
- \( n₁ \) = moles of solvent (A)
- \( n₂ \) = moles of solute (B)

Substituting the values:
\[ \frac{10 - 9}{10} = \frac{n₂}{n₁ + n₂} \]
\[ \frac{1}{10} = \frac{n₂}{n₁ + n₂} \]

### Step 3: Calculate moles of solvent (A)
The number of moles of A (n₁) can be calculated using:
\[ n₁ = \frac{m₁}{M₁} = \frac{20 \text{ g}}{200 \text{ g/mol}} = 0.1 \text{ mol} \]

### Step 4: Set up the equation
From the previous step, we have:
\[ \frac{1}{10} = \frac{n₂}{0.1 + n₂} \]

### Step 5: Cross-multiply to solve for n₂
Cross-multiplying gives:
\[ 1 \cdot (0.1 + n₂) = 10 \cdot n₂ \]
\[ 0.1 + n₂ = 10n₂ \]
\[ 0.1 = 10n₂ - n₂ \]
\[ 0.1 = 9n₂ \]
\[ n₂ = \frac{0.1}{9} \approx 0.0111 \text{ mol} \]

### Step 6: Calculate the molecular mass of B
Now we can find the molecular mass (M₂) of B using:
\[ n₂ = \frac{m₂}{M₂} \]
Rearranging gives:
\[ M₂ = \frac{m₂}{n₂} = \frac{1 \text{ g}}{0.0111 \text{ mol}} \approx 90 \text{ g/mol} \]

### Final Answer
The molecular mass of B is approximately **90 amu**.

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