In a hypothetical reaction
`A(aq)hArr2B(aq)+C(aq)` (`1^(st)` order decomposition)
'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with `H_2O_2`.Hence the progress of reaction can be monitored by measuring rotation of plane of polarised light or by measuring volume of `H_2O_2` consumed in titration.
In an experiment, the optical rotation was found to be `theta=30^@C` at t=20 min. and `theta=15^(@)` at t=50min from start of the reaction.If the progress would have been monitored by titration method, volume of `H_2O_2` consumed at t=30 min.(from start) is 30 ml then volume of `H_2O_2` consumed at t=90 min, will be:

Options

A. 60 ml
B. 45 ml
C. 52.5 ml
D. 90 ml

Explanation

To solve the problem, we need to determine the volume of H₂O₂ consumed at t = 90 minutes based on the information given about the reaction and the optical rotation measurements.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction is: A(aq) ⇌ 2B(aq) + C(aq)
- A is optically active (dextrorotatory), while B and C are optically inactive.
- The progress of the reaction can be monitored through optical rotation or by measuring the volume of H₂O₂ consumed in a titration reaction with B.

2. **Optical Rotation Measurements**:
- At t = 20 minutes, the optical rotation (θ) is 30°.
- At t = 50 minutes, the optical rotation (θ) is 15°.
- The decrease in optical rotation indicates the consumption of A, which is optically active.

3. **Calculating the Half-Life**:
- The change in optical rotation from 30° to 15° suggests that the concentration of A has decreased to half its initial value in the time interval from 20 to 50 minutes.
- The time difference between these two measurements is 30 minutes. Hence, the half-life (t₁/₂) of the reaction is 30 minutes.

4. **Volume of H₂O₂ Consumed**:
- It is given that at t = 30 minutes, the volume of H₂O₂ consumed is 30 mL.
- Since the reaction is first-order, we can use the half-life to find the volumes consumed at subsequent half-lives.

5. **Calculating H₂O₂ Consumption at t = 90 minutes**:
- From t = 30 minutes to t = 90 minutes, there are two additional half-lives (30 minutes each).
- Therefore, we will have:
- At t = 30 minutes: 30 mL of H₂O₂ consumed.
- At t = 60 minutes (1 half-life later): 30 mL / 2 = 15 mL consumed.
- At t = 90 minutes (2 half-lives later): 15 mL / 2 = 7.5 mL consumed.

6. **Total Volume of H₂O₂ Consumed at t = 90 minutes**:
- Total volume consumed = Volume at t = 30 minutes + Volume at t = 60 minutes + Volume at t = 90 minutes
- Total volume = 30 mL + 15 mL + 7.5 mL = 52.5 mL.

### Final Answer:
The volume of H₂O₂ consumed at t = 90 minutes is **52.5 mL**.