X gm of `KHC_(2)O_(4)` requires 100 ml of `0.02 M KMnO_(4)` in acidic medium. In another experiment, y gm of `KHC_(2)O_(4)` requires 100 mkl of 0.05 M `Ca(OH)_(2)`. The ratio of x and y is

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `5:4`

Answer

To solve the problem, we need to determine the ratio of \( x \) and \( y \) where \( x \) grams of \( KHC_2O_4 \) reacts with \( KMnO_4 \) and \( y \) grams of \( KHC_2O_4 \) reacts with \( Ca(OH)_2 \).

### Step 1: Calculate the equivalents of \( KHC_2O_4 \) required for \( KMnO_4 \)

1. **Given Data**:
- Volume of \( KMnO_4 \) solution = 100 mL = 0.1 L
- Molarity of \( KMnO_4 \) = 0.02 M

2. **Calculate moles of \( KMnO_4 \)**:
\[ \text{Moles of } KMnO_4 = \text{Molarity} \times \text{Volume} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} \]

3. **Determine the equivalents of \( KMnO_4 \)**:
- The reaction of \( KMnO_4 \) in acidic medium shows that 1 mole of \( KMnO_4 \) provides 5 equivalents.
\[ \text{Equivalents of } KMnO_4 = 0.002 \, \text{mol} \times 5 = 0.01 \, \text{equivalents} \]

4. **Relate it to \( KHC_2O_4 \)**:
- Let the equivalent weight of \( KHC_2O_4 \) be \( \frac{M}{1} \) (where \( M \) is the molar mass of \( KHC_2O_4 \)).
- Therefore, \( \frac{x}{M} = 0.01 \) (since 1 equivalent of \( KHC_2O_4 \) reacts with 1 equivalent of \( KMnO_4 \)).
\[ x = 0.01 \times M \]

### Step 2: Calculate the equivalents of \( KHC_2O_4 \) required for \( Ca(OH)_2 \)

1. **Given Data**:
- Volume of \( Ca(OH)_2 \) solution = 100 mL = 0.1 L
- Molarity of \( Ca(OH)_2 \) = 0.05 M

2. **Calculate moles of \( Ca(OH)_2 \)**:
\[ \text{Moles of } Ca(OH)_2 = 0.05 \, \text{mol/L} \times 0.1 \, \text{L} = 0.005 \, \text{mol} \]

3. **Determine the equivalents of \( Ca(OH)_2 \)**:
- Each mole of \( Ca(OH)_2 \) provides 2 equivalents.
\[ \text{Equivalents of } Ca(OH)_2 = 0.005 \, \text{mol} \times 2 = 0.01 \, \text{equivalents} \]

4. **Relate it to \( KHC_2O_4 \)**:
- Therefore, \( \frac{y}{M} = 0.01 \).
\[ y = 0.01 \times M \]

### Step 3: Find the ratio \( \frac{x}{y} \)

1. **Substituting the values of \( x \) and \( y \)**:
\[ \frac{x}{y} = \frac{0.01 \times M}{0.01 \times M} = 1 \]

2. **Thus, the ratio \( x:y \)**:
\[ x:y = 1:1 \]

### Conclusion

The ratio of \( x \) and \( y \) is \( 1:1 \).

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