An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the value of t1/8/t1/10 × 10?

For a first-order reaction, the integrated rate law is given by:

ln(A) = -kt + ln(A₀)

where:

• A is the concentration at time t
• A₀ is the initial concentration
• k is the rate constant

We can rewrite this as:

ln(A/A₀) = -kt

For t1/8, A/A₀ = 1/8:

ln(1/8) = -kt1/8

3ln(1/2) = -kt1/8

-3ln(2) = -kt1/8

For t1/10, A/A₀ = 1/10:

ln(1/10) = -kt1/10

-ln(10) = -kt1/10

Now we can find the ratio t1/8/t1/10:

t1/8/t1/10 = (3ln2)/(ln10)

Using a calculator:

ln(2) ≈ 0.693

ln(10) ≈ 2.303

Therefore:

t1/8/t1/10 ≈ (3 * 0.693) / 2.303 ≈ 2.079 / 2.303 ≈ 0.9028

Now, we need to find the value of t1/8/t1/10 × 10:

(t1/8/t1/10) × 10 ≈ 0.9028 × 10 ≈ 9

Therefore, the closest answer is 9.