STATEMENT-1 : In the balanced redox reaction,
`x As_(2)S_(3)+y NO_(3)^(-)+4H_(2)Orarr aAsO_(4)^(3-)+bNO+cSO_(4)^(2-)+8H^(+)` the n-factor of `As_(2)S_(3)` and `NO_(3)^(-)` is 28 and 3 respectively.
Statement-2 : Molar ratio is reciprocal of n-factor's ratio so `x : t` is `3 : 28`.

Options

A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT-1
B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
C. If STATEMENT- is 1 TRUE and STATEMENT-2 is FALSE
D. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer

To solve the problem, we need to analyze the statements regarding the balanced redox reaction and the n-factors of the reactants involved.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The balanced redox reaction is given as:
\[ x \, \text{As}_2\text{S}_3 + y \, \text{NO}_3^{-} + 4 \, \text{H}_2\text{O} \rightarrow a \, \text{AsO}_4^{3-} + b \, \text{NO} + c \, \text{SO}_4^{2-} + 8 \, \text{H}^{+} \]

2. **Determine the n-factor of As\(_2\)S\(_3\)**:
- The oxidation state of As in As\(_2\)S\(_3\) is calculated. Sulfur (S) has an oxidation state of -2, so:
\[ 2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies x = +3 \]
- In the product AsO\(_4^{3-}\), the oxidation state of As is +5.
- The change in oxidation state for each As atom is from +3 to +5 (2 electrons lost per As atom). Since there are 2 As atoms, the total change is:
\[ 2 \text{ electrons} \times 2 = 4 \text{ electrons lost} \]
- Additionally, S in As\(_2\)S\(_3\) is converted to SO\(_4^{2-}\) (oxidation state +6). The change for S is from -2 to +6 (8 electrons lost).
- Therefore, for 2 S atoms, the total change is:
\[ 8 \text{ electrons} \times 2 = 16 \text{ electrons lost} \]
- Total n-factor for As\(_2\)S\(_3\):
\[ 4 + 16 = 20 \text{ electrons} \]

3. **Determine the n-factor of NO\(_3^{-}\)**:
- NO\(_3^{-}\) is reduced to NO. The oxidation states are:
- In NO\(_3^{-}\), N is +5.
- In NO, N is +2.
- The change in oxidation state is from +5 to +2 (3 electrons gained).
- Thus, the n-factor for NO\(_3^{-}\) is 3.

4. **Establish the Relationship Between n-factors and Molar Ratios**:
- The n-factor of As\(_2\)S\(_3\) is 28 and for NO\(_3^{-}\) is 3.
- The molar ratio is the reciprocal of the n-factor ratio:
\[ \frac{x}{y} = \frac{n_{NO_3^{-}}}{n_{As_2S_3}} = \frac{3}{28} \]

5. **Conclusion**:
- The statement that the molar ratio \(x : y\) is \(3 : 28\) is correct based on the n-factors calculated.

### Final Answer:
The correct answer is that the molar ratio \(x : y\) is indeed \(3 : 28\).

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