A current is passed through two cells connected in series. The first cell contains X`(NO_(3) )_( 3)` (aq) and the second cell contains Y`(NO_(3) )_( 2)` (aq). The relative atomic masses of X and Y are in the ratio `1: 2`. What is the ratio of liberated mass of X to that of Y?

Options

A. `3 : 2`
B. `1:2`
C. `1:3`
D. `3:1`

Answer

To solve the problem, we need to find the ratio of the liberated mass of X to that of Y when a current is passed through two electrochemical cells connected in series.

### Step-by-Step Solution:

1. **Identify the Atomic Masses**:
- Let the atomic mass of X be \( A \).
- Given that the atomic mass of Y is in the ratio of 1:2 with X, we can express the atomic mass of Y as \( 2A \).

2. **Determine the Equivalent Weights**:
- The equivalent weight is defined as the atomic mass divided by the valency.
- For X, which is in the form \( X(NO_3)_3 \), the valency is 3 (since it can donate 3 electrons). Thus, the equivalent weight of X is:
\[ \text{Equivalent weight of X} = \frac{A}{3} \]
- For Y, which is in the form \( Y(NO_3)_2 \), the valency is 2 (since it can donate 2 electrons). Thus, the equivalent weight of Y is:
\[ \text{Equivalent weight of Y} = \frac{2A}{2} = A \]

3. **Establish the Relationship Between Masses**:
- The mass of a substance liberated during electrolysis is directly proportional to its equivalent weight. Therefore, we can write:
\[ \frac{\text{Mass of X}}{\text{Mass of Y}} = \frac{\text{Equivalent weight of X}}{\text{Equivalent weight of Y}} \]
- Substituting the equivalent weights we calculated:
\[ \frac{\text{Mass of X}}{\text{Mass of Y}} = \frac{\frac{A}{3}}{A} \]

4. **Simplify the Ratio**:
- Simplifying the right-hand side:
\[ \frac{\text{Mass of X}}{\text{Mass of Y}} = \frac{A/3}{A} = \frac{1}{3} \]

5. **Conclusion**:
- The ratio of the liberated mass of X to that of Y is:
\[ \text{Mass of X} : \text{Mass of Y} = 1 : 3 \]

### Final Answer:
The ratio of the liberated mass of X to that of Y is \( 1 : 3 \).