A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are $$300\,\text{s}$$ and $$180\,\text{s}$$, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is: (Use $$\ln 2 = 0.693$$)

For a first-order reaction, the half-life $$t_{1/2}$$ is related to the rate constant $$k$$ by the formula

$$t_{1/2}=\frac{0.693}{k}\,.$$

We have the following data:

For compound A, $$t_{1/2(A)} = 300\,\text{s}$$, so

$$k_A = \frac{0.693}{t_{1/2(A)}} = \frac{0.693}{300} = 0.00231\ \text{s}^{-1}.$$

For compound B, $$t_{1/2(B)} = 180\,\text{s}$$, so

$$k_B = \frac{0.693}{t_{1/2(B)}} = \frac{0.693}{180} = 0.00385\ \text{s}^{-1}.$$

Let the initial concentrations of A and B be equal and denoted by $$[A]_0 = [B]_0$$. For first-order kinetics, the concentration at time $$t$$ is

$$[A] = [A]_0 e^{-k_A t}, \qquad [B] = [B]_0 e^{-k_B t}\,.$$

We are asked to find the time $$t$$ when the concentration of A becomes four times that of B, i.e.

$$[A] = 4[B].$$

Substituting the exponential expressions, we get

$$[A]_0 e^{-k_A t} = 4\,[B]_0 e^{-k_B t}.$$

Because $$[A]_0 = [B]_0$$, these cancel, leaving

$$e^{-k_A t} = 4\,e^{-k_B t}.$$

Dividing both sides by $$e^{-k_B t}$$ gives

$$e^{\,(-k_A + k_B)\,t} = 4.$$

Taking natural logarithms on both sides,

$$( -k_A + k_B )\,t = \ln 4.$$

Since $$\ln 4 = 2\ln 2 = 2 \times 0.693 = 1.386$$, we have

$$t = \frac{1.386}{k_B - k_A}.$$

Now substitute the calculated rate constants:

$$k_B - k_A = 0.00385 - 0.00231 = 0.00154\ \text{s}^{-1},$$

so

$$t = \frac{1.386}{0.00154} = 900\ \text{s}.$$

Hence, the correct answer is Option B.