15 gm `Ba(MnO_4)_2` sample containing inert impurity is completely reacting with 100 ml of 11.2 V' `H_2O_2`, then what will be the % purity of `Ba(MnO_4)_2` in the sample ?
(Atomic mass Ba=137, Mn=55)

Options

A. `5%`
B. `10%`
C. `50%`
D. none

Explanation

To solve the problem of determining the percentage purity of `Ba(MnO_4)_2` in a 15 g sample that reacts completely with 100 ml of 11.2 volume `H_2O_2`, we can follow these steps:

### Step 1: Calculate the Normality of `H_2O_2`
The volume of `H_2O_2` is given as 11.2 volumes, which means it can release 11.2 volumes of oxygen gas. The normality (N) of `H_2O_2` can be calculated as:
\[ \text{Normality} = \frac{\text{Volume}}{5.6} \]
\[ \text{Normality} = \frac{11.2}{5.6} = 2 \, \text{N} \]

### Step 2: Calculate Milliequivalents of `H_2O_2`
To find the milliequivalents of `H_2O_2` in 100 ml:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (L)} \]
\[ \text{Milliequivalents} = 2 \, \text{N} \times 0.1 \, \text{L} = 0.2 \, \text{Eq} = 200 \, \text{mEq} \]

### Step 3: Determine the Milliequivalents of `Ba(MnO_4)_2`
The reaction of `Ba(MnO_4)_2` with `H_2O_2` involves a stoichiometry where 1 mole of `Ba(MnO_4)_2` gives 10 equivalents (since it has 2 MnO₄⁻ ions, each contributing 5 equivalents). Therefore:
\[ \text{Milliequivalents of } Ba(MnO_4)_2 = \frac{200 \, \text{mEq}}{10} = 20 \, \text{mEq} \]

### Step 4: Calculate the Number of Moles of `Ba(MnO_4)_2`
Using the relationship between milliequivalents and moles:
\[ \text{Moles} = \frac{\text{Milliequivalents}}{n \text{ factor}} \]
Where the n factor for `Ba(MnO_4)_2` is 10 (as calculated above):
\[ \text{Moles of } Ba(MnO_4)_2 = \frac{20 \, \text{mEq}}{10} = 2 \, \text{moles} \]

### Step 5: Calculate the Molar Mass of `Ba(MnO_4)_2`
The molar mass of `Ba(MnO_4)_2` can be calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of Mn = 55 g/mol
- Molar mass of O = 16 g/mol

Thus,
\[ \text{Molar mass of } Ba(MnO_4)_2 = 137 + 2 \times (55 + 4 \times 16) \]
\[ = 137 + 2 \times (55 + 64) \]
\[ = 137 + 2 \times 119 \]
\[ = 137 + 238 = 375 \, \text{g/mol} \]

### Step 6: Calculate the Weight of `Ba(MnO_4)_2` in the Sample
Now, we can find the weight of `Ba(MnO_4)_2` in the sample:
\[ \text{Weight of } Ba(MnO_4)_2 = \text{Moles} \times \text{Molar Mass} \]
\[ = 2 \, \text{moles} \times 375 \, \text{g/mol} = 750 \, \text{g} \]

### Step 7: Calculate the Percentage Purity
Finally, we calculate the percentage purity of `Ba(MnO_4)_2` in the original sample:
\[ \text{Percentage Purity} = \left( \frac{\text{Weight of } Ba(MnO_4)_2}{\text{Total Sample Weight}} \right) \times 100 \]
\[ = \left( \frac{750 \, \text{g}}{15 \, \text{g}} \right) \times 100 = 50\% \]

### Final Answer
The percentage purity of `Ba(MnO_4)_2` in the sample is **50%**.

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