Equal mass of a soute are dissolved in equal mass of two solvents A and B and formed very dilute solution. The relative lowering of vapour pressure for the solution B has twice the relative lowering of vapour pressure for the solution A. If `m_(A) "and" M_(B)` are the molecules mass of solventds A and B respectively, then :

Options

A. `M_(A) =M_(B)`
B. `M_(B) =2xxM_(A)`
C. `M_(A) =4M_(B)`
D. `M_(A) =2M_(B)`

Answer

To solve the problem step by step, we will analyze the information given and apply the relevant concepts in chemistry.

### Step 1: Understand the Concept of Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure is defined as:
\[ \text{Relative Lowering of Vapor Pressure} = \frac{P_0 - P_s}{P_0} \]
where \(P_0\) is the vapor pressure of the pure solvent and \(P_s\) is the vapor pressure of the solution. According to Raoult's law, this relative lowering is proportional to the mole fraction of the solute in the solution.

### Step 2: Set Up the Ratios for the Two Solutions
We are given that the relative lowering of vapor pressure for solution B is twice that of solution A:
\[ \text{Relative Lowering of Vapor Pressure for B} = 2 \times \text{Relative Lowering of Vapor Pressure for A} \]

### Step 3: Express the Relative Lowering in Terms of Moles
Using the definition of relative lowering:
\[ \frac{P_0 - P_{sA}}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_A} \quad \text{(for solution A)} \]
\[ \frac{P_0 - P_{sB}}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_B} \quad \text{(for solution B)} \]

### Step 4: Simplify for Very Dilute Solutions
For very dilute solutions, the number of moles of solute is much smaller than that of the solvent, allowing us to approximate:
\[ \frac{P_0 - P_{sA}}{P_0} \approx \frac{n_{\text{solute}}}{n_A} \]
\[ \frac{P_0 - P_{sB}}{P_0} \approx \frac{n_{\text{solute}}}{n_B} \]

### Step 5: Set Up the Equation Based on Given Information
From the problem statement, we can write:
\[ \frac{n_B}{n_A} = 2 \]

### Step 6: Relate Moles to Mass and Molecular Mass
The number of moles is given by:
\[ n = \frac{m}{M} \]
where \(m\) is the mass and \(M\) is the molecular mass. Since equal mass of solute is dissolved in equal mass of solvents:
\[ n_A = \frac{m_A}{M_A} \quad \text{and} \quad n_B = \frac{m_B}{M_B} \]
Given that \(m_A = m_B\), we can denote this mass as \(m\):
\[ n_A = \frac{m}{M_A} \quad \text{and} \quad n_B = \frac{m}{M_B} \]

### Step 7: Substitute into the Ratio
Substituting into the ratio gives:
\[ \frac{n_B}{n_A} = \frac{\frac{m}{M_B}}{\frac{m}{M_A}} = \frac{M_A}{M_B} \]
Setting this equal to 2 (from step 5):
\[ \frac{M_A}{M_B} = 2 \implies M_B = \frac{M_A}{2} \]

### Step 8: Conclusion
Thus, we find the relationship between the molecular masses of the solvents:
\[ M_B = 2 \times M_A \]

### Final Answer
The correct relation between the molecular masses of the solvents A and B is:
\[ M_B = 2 \times M_A \]