Out of300g substance [decomposes as per 1st order]. How much (nearly)will remain after 18 hr?`(t_(1//2) = 3hr)`

Options

A. 4.6 gm
B. 5.6 gm
C. 9.2 gm
D. 6.4 gm

Explanation

To solve the problem of how much of the 300 g substance will remain after 18 hours, given that it decomposes according to first-order kinetics with a half-life of 3 hours, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the rate constant (k)**:
The formula for the half-life (t₁/₂) of a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Rearranging this to find k:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3 \text{ hours}} \approx 0.231 \text{ hr}^{-1} \]

2. **Calculate the number of half-lives in 18 hours**:
Since the half-life is 3 hours, we can find out how many half-lives fit into 18 hours:
\[ \text{Number of half-lives} = \frac{18 \text{ hours}}{3 \text{ hours}} = 6 \]

3. **Calculate the remaining amount using the half-life formula**:
The amount remaining after n half-lives can be calculated using the formula:
\[ A = A_0 \left(\frac{1}{2}\right)^n \]
Where:
- \( A_0 = 300 \text{ g} \)
- \( n = 6 \)
Substituting the values:
\[ A = 300 \left(\frac{1}{2}\right)^6 = 300 \left(\frac{1}{64}\right) = \frac{300}{64} \approx 4.6875 \text{ g} \]

4. **Round the final answer**:
Rounding 4.6875 g gives approximately 4.7 g. However, since the question asks for "nearly," we can state that about 4.6 g will remain.

### Final Answer:
Approximately **4.6 g** of the substance will remain after 18 hours.