Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

Options

A. 20
B. 50
C. 100
D. 200

Explanation

To solve the problem step by step, we will use the concept of vapor pressure lowering and the relationship between the number of moles of solute and solvent.

### Step 1: Identify Given Values
- Vapor pressure of pure liquid X (P0) = 2 atm
- Vapor pressure of the solution (Ps) = 1 atm
- Mass of solute Y (wB) = 1 g
- Mass of solvent X (wA) = 20 g
- Molar mass of X (mA) = 200 g/mol

### Step 2: Calculate the Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure can be calculated using the formula:
\[ \text{Relative lowering} = \frac{P_0 - P_s}{P_0} = \frac{2 - 1}{2} = \frac{1}{2} \]

### Step 3: Calculate the Number of Moles of Solvent (X)
The number of moles of solvent (X) can be calculated using the formula:
\[ n_A = \frac{w_A}{m_A} \]
Substituting the values:
\[ n_A = \frac{20 \, \text{g}}{200 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 4: Set Up the Equation for the Number of Moles of Solute (Y)
Let the molar mass of Y be \( m_B \). The number of moles of solute (Y) is given by:
\[ n_B = \frac{w_B}{m_B} = \frac{1 \, \text{g}}{m_B} \]

### Step 5: Use the Dilute Solution Approximation
For a dilute solution, we can use the approximation:
\[ \frac{n_B}{n_A} \approx \frac{P_0 - P_s}{P_s} \]
Substituting the values we have:
\[ \frac{1/m_B}{0.1} = \frac{1/2} \]

### Step 6: Solve for Molar Mass of Y (mB)
Rearranging the equation gives:
\[ \frac{1}{m_B} = \frac{0.1}{2} \]
Thus,
\[ m_B = \frac{0.1 \times 2}{1} = 0.2 \, \text{g/mol} \]

### Step 7: Convert to g/mol
Since we need the molar mass in g/mol, we multiply by 1000 to convert from kg to g:
\[ m_B = 20 \, \text{g/mol} \]

### Final Answer
The molar mass of Y is **20 g/mol**.

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