The strength of `H_(2)O_(2)` is expressed in many ways like molarity , normality , % strength and volume strengths But out of all these form of strengths , volume strength has great significance for chemical reactions .
This decomposition of `H_(2)O_(2)` is shown as under
`H_(2)O_(2)(l) to H_(2)O(l) + (1)/(2)O_(2)(g)`
'x' volume strength of `H_(2)O_(2)` means one volume (litre or ml) of `H_(2)O_(2)` releases x volume (litre or ml) of `O_(2)` at NTP .
1 litre `H_(2)O_(2)` release x litre of `O_(2)` at NTP
`=(x)/(22.4)` moles of `O_(2)`
From the equation ,
1 mole of `O_(2)` produces from 2 moles of `H_(2)O_(2)` .
`(x)/(22.4)` moles of `O_(2)` produces from `2xx(x)/(22.4)` moles of `H_(2)O_(2)`
`=(x)/(11.2)` moles of `H_(2)O_(2)`
So, molarity of `H_(2)O_(2)=((x)/(11.2))/(1)=(x)/(11.2)M`
Normality =n-factor `xx` molarity
`=2xx(x)/(11.2)=(x)/(5.6) N`
30 g `Ba(MnO_(4))_(2)` sample containing inert impurity is completely reacting with 100 ml of "28 volume" strength of `H_(2)O_(2)` in acidic medium then what will be the percentage purity of `Ba(MnO_(4))_(2)` in the sample ? (Ba=137, Mn=55, O=16)

Options

A. `10%`
B. `40%`
C. `62.5%`
D. `80%`

Answer

To solve the problem, we will go through the following steps:

### Step 1: Calculate the Molarity of H₂O₂
Given that the volume strength of H₂O₂ is 28, we can find the molarity using the relationship derived in the transcript:

\[ \text{Molarity of } H_2O_2 = \frac{x}{11.2} \text{ M} \]

Here, \( x = 28 \):

\[ \text{Molarity of } H_2O_2 = \frac{28}{11.2} = 2.5 \text{ M} \]

### Step 2: Calculate the Normality of H₂O₂
The normality (N) is calculated using the formula:

\[ \text{Normality} = n \text{-factor} \times \text{Molarity} \]

The n-factor for H₂O₂ (as it loses 2 electrons) is 2:

\[ \text{Normality of } H_2O_2 = 2 \times 2.5 = 5 \text{ N} \]

### Step 3: Calculate the Equivalents of H₂O₂
The volume of H₂O₂ used is 100 mL, which is 0.1 L. The equivalents of H₂O₂ can be calculated as follows:

\[ \text{Equivalents of } H_2O_2 = \text{Normality} \times \text{Volume in L} \]

\[ \text{Equivalents of } H_2O_2 = 5 \times 0.1 = 0.5 \text{ equivalents} \]

### Step 4: Relate Equivalents of H₂O₂ to Ba(MnO₄)₂
The reaction shows that 1 equivalent of H₂O₂ reacts with 1 equivalent of Ba(MnO₄)₂. Therefore, the equivalents of Ba(MnO₄)₂ will also be 0.5 equivalents.

### Step 5: Calculate the Molar Mass of Ba(MnO₄)₂
The molar mass of Ba(MnO₄)₂ can be calculated as follows:

\[ \text{Molar mass of Ba(MnO₄)₂} = 137 + 2 \times 55 + 8 \times 16 = 375 \text{ g/mol} \]

### Step 6: Calculate the Mass of Ba(MnO₄)₂
Using the equivalents calculated earlier, we can find the mass of Ba(MnO₄)₂ present in the sample:

\[ \text{Mass of Ba(MnO₄)₂} = \text{Equivalents} \times \text{Molar mass} = 0.5 \times 375 = 187.5 \text{ g} \]

### Step 7: Calculate the Percentage Purity of Ba(MnO₄)₂
The percentage purity can be calculated using the formula:

\[ \text{Percentage Purity} = \left( \frac{\text{Mass of Ba(MnO₄)₂}}{\text{Total mass of sample}} \right) \times 100 \]

Given the total mass of the sample is 30 g:

\[ \text{Percentage Purity} = \left( \frac{187.5}{30} \right) \times 100 = 62.5\% \]

### Final Answer
The percentage purity of Ba(MnO₄)₂ in the sample is **62.5%**.

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