A current is passed through 2 voltmeters connected in series. The first voltmeter contians `XSO_4)(aq)` and second has `Y_(2)SO_(4)`(aq). The relative atomic masses of X and Y are in the ratio `2:1`. The ratio of the mass of X liberated to the mass of Y liberated is:

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. none of above

Explanation

To solve the problem, we need to determine the ratio of the mass of X liberated to the mass of Y liberated when a current is passed through two voltmeters containing different sulfate solutions.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two voltmeters in series: the first contains \( XSO_4(aq) \) and the second contains \( Y_2SO_4(aq) \).
- The relative atomic masses of elements X and Y are in the ratio \( 2:1 \).

2. **Let’s Define the Variables**:
- Let the relative atomic mass of Y be \( m \). Therefore, the relative atomic mass of X will be \( 2m \).

3. **Charge Passed**:
- Assume a current of \( 1 \) ampere is passed for \( 1 \) second. The total charge \( Q \) passed is:
\[ Q = 1 \, \text{A} \times 1 \, \text{s} = 1 \, \text{C} \]

4. **Calculating Moles of Electrons**:
- The number of moles of electrons transferred can be calculated using Faraday's constant (\( F = 96500 \, \text{C/mol} \)):
\[ \text{Moles of electrons} = \frac{Q}{F} = \frac{1 \, \text{C}}{96500 \, \text{C/mol}} = \frac{1}{96500} \, \text{mol} \]

5. **Mass of X Liberated**:
- In \( XSO_4 \), X is in the +2 oxidation state. Thus, it requires \( 2 \) moles of electrons to liberate \( 1 \) mole of X:
\[ \text{Moles of X liberated} = \frac{1}{2} \times \frac{1}{96500} = \frac{1}{193000} \, \text{mol} \]
- The mass of X liberated is:
\[ \text{Mass of X} = \text{Moles of X} \times \text{Molar mass of X} = \frac{1}{193000} \times 2m = \frac{2m}{193000} \, \text{g} \]

6. **Mass of Y Liberated**:
- In \( Y_2SO_4 \), Y is in the +2 oxidation state. Thus, it requires \( 1 \) mole of electrons to liberate \( 1 \) mole of Y:
\[ \text{Moles of Y liberated} = \frac{1}{96500} \, \text{mol} \]
- The mass of Y liberated is:
\[ \text{Mass of Y} = \text{Moles of Y} \times \text{Molar mass of Y} = \frac{1}{96500} \times m = \frac{m}{96500} \, \text{g} \]

7. **Finding the Ratio of Masses**:
- Now, we can find the ratio of the mass of X liberated to the mass of Y liberated:
\[ \text{Ratio} = \frac{\text{Mass of X}}{\text{Mass of Y}} = \frac{\frac{2m}{193000}}{\frac{m}{96500}} = \frac{2m \cdot 96500}{m \cdot 193000} \]
- Simplifying this gives:
\[ \text{Ratio} = \frac{2 \cdot 96500}{193000} = \frac{193000}{193000} = 1 \]

### Final Answer:
The ratio of the mass of X liberated to the mass of Y liberated is \( 1:1 \).

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