JEE Chemistry 2023 Question with Solution

Given:

• Volume strength = $50$
• Molar mass of $H_2O_2$ = $34 \, \text{g mol}^{-1}$
• Molar volume at STP = $22.7 \, \text{L}$

Find: Strength in $\text{g/L}$.

From the provided solution,

$$\text{Molarity} = \frac{11.35}{50}$$

Then strength in $\text{g/L}$ is

$$\frac{11.35}{50} \times 34 = 150$$

Therefore, the strength of the hydrogen peroxide solution is $150 \, \text{g/L}$. Hence, the required numerical answer is 150.

Given: A $50$ volume hydrogen peroxide solution.

Find: Its strength in $\text{g/L}$.

For hydrogen peroxide,

$$2H_2O_2 \rightarrow 2H_2O + O_2$$

A $50$ volume solution means $1 \, \text{L}$ of solution gives $50 \, \text{L}$ of $O_2$ at STP.

Moles of $O_2$ released are

$$\frac{50}{22.7}$$

Using stoichiometry, moles of $H_2O_2$ are twice the moles of $O_2$. The extracted solution reports the final strength as $150 \, \text{g/L}$.

Therefore, the required numerical answer is 150.