The percentage labelling (mixture of `H_(2)SO_(4)` and `SO_(3))` refers to the total mass of pure `H_(2)SO_(4)`. The total amount of `H_(2)SO_(4)` found after adding calculated amount of water to `100 g` oleum is the percentage labelling of oleum.Higher the percentage lebeling of oleum higher is the amount of free `SO_(3)` in the oleum sample.
`100 g` sample of '149%' oleum was taken and calculated amount of `H_(2)O` was added to make `H_(2)SO_(4)`.`500 mL` solution of `x MKOH` solution is required to neutralize the solution. The value of `x` is.

Options

A. `1 M`
B. `2 M`
C. `4 M`
D. `6 M`

Explanation

To solve the problem step-by-step, we need to determine how much KOH is required to neutralize the H2SO4 produced from the oleum. Here’s the step-by-step solution:

### Step 1: Understand the composition of 149% oleum
- The percentage labeling of oleum indicates that for every 100 grams of oleum, there are 149 grams of H2SO4.
- This means that in 100 grams of 149% oleum, we have:
- Mass of H2SO4 = 100 g (pure)
- Mass of SO3 = 49 g (since 149% - 100% = 49%)

### Step 2: Calculate the total mass of H2SO4 after adding water
- When water is added to oleum, the SO3 reacts with water to form more H2SO4.
- The reaction is: SO3 + H2O → H2SO4
- Therefore, the total mass of H2SO4 after adding water is:
- Total H2SO4 = 100 g (from oleum) + 49 g (from SO3) = 149 g

### Step 3: Calculate the number of equivalents of H2SO4
- The equivalent mass of H2SO4 can be calculated using its molar mass and basicity (n-factor).
- Molar mass of H2SO4 = 98 g/mol
- Basicity of H2SO4 = 2 (it can donate 2 H+ ions)
- Equivalent mass of H2SO4 = Molar mass / n-factor = 98 g / 2 = 49 g/equiv
- Number of equivalents of H2SO4 = Mass of H2SO4 / Equivalent mass = 149 g / 49 g/equiv = 3.04 equiv

### Step 4: Calculate the number of equivalents of KOH required
- KOH is a strong base and will react with H2SO4 in a 1:1 ratio.
- Therefore, the number of equivalents of KOH required will be equal to the number of equivalents of H2SO4, which is 3.04 equiv.

### Step 5: Relate KOH equivalents to molarity and volume
- The number of equivalents of KOH can also be calculated using its molarity (x) and volume (500 mL).
- Number of equivalents of KOH = Molarity × Volume (in L)
- Volume in liters = 500 mL = 0.5 L
- Therefore, equivalents of KOH = x × 0.5

### Step 6: Set up the equation
- Since the number of equivalents of KOH equals the number of equivalents of H2SO4:
\[ x \times 0.5 = 3.04 \]

### Step 7: Solve for x
- Rearranging the equation gives:
\[ x = \frac{3.04}{0.5} = 6.08 \, \text{M} \]
- Rounding off, we find:
\[ x \approx 6 \, \text{M} \]

### Conclusion
The value of x, which represents the molarity of the KOH solution required to neutralize the H2SO4 produced from the oleum, is **6 M**.

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