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`

Answer

To solve the problem, we will use Raoult's Law and the information given in the question. Let's break it down step by step.

### Step 1: Understand the Given Data
- The vapor pressure of pure liquid A, \( P_0 = 10 \) torr.
- The vapor pressure of the solution, \( P = 9 \) torr.
- Mass of A, \( m_A = 20 \) g.
- Molecular mass of A, \( M_A = 200 \) g/mol.
- Mass of B, \( m_B = 1 \) g.
- We need to find the molecular mass of B, \( M_B \).

### Step 2: Calculate the Moles of A
Using the formula for moles:
\[ n_A = \frac{m_A}{M_A} \]
Substituting the values:
\[ n_A = \frac{20 \, \text{g}}{200 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Use Raoult's Law
Raoult's Law states:
\[ \frac{P_0 - P}{P_0} = \frac{n_B}{n_A + n_B} \]
Where \( n_B \) is the number of moles of B.

### Step 4: Calculate the Change in Vapor Pressure
Substituting the values into Raoult's Law:
\[ \frac{10 - 9}{10} = \frac{n_B}{n_A + n_B} \]
This simplifies to:
\[ \frac{1}{10} = \frac{n_B}{n_A + n_B} \]

### Step 5: Express Moles of B
We know that:
\[ n_B = \frac{m_B}{M_B} = \frac{1 \, \text{g}}{M_B} \]
Substituting \( n_B \) into the equation:
\[ \frac{1}{10} = \frac{\frac{1}{M_B}}{0.1 + \frac{1}{M_B}} \]

### Step 6: Cross Multiply and Simplify
Cross-multiplying gives:
\[ 1 \cdot (0.1 + \frac{1}{M_B}) = 10 \cdot \frac{1}{M_B} \]
This simplifies to:
\[ 0.1 + \frac{1}{M_B} = \frac{10}{M_B} \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 0.1 = \frac{10}{M_B} - \frac{1}{M_B} \]
\[ 0.1 = \frac{10 - 1}{M_B} \]
\[ 0.1 = \frac{9}{M_B} \]

### Step 8: Solve for \( M_B \)
Now, solving for \( M_B \):
\[ M_B = \frac{9}{0.1} = 90 \, \text{g/mol} \]

### Conclusion
The molecular mass of B is \( 90 \, \text{g/mol} \).