Question

Decomposition of $H_2{O}_2$ is a first order reaction. Initially, a solution of $H_2{O}_2$ having half-life time $15$ min has a strength of $16$ volume. Time taken for a solution to become $1$ volume is:

• A. $4$ min
• B. $15$ min
• C. $30$ min
• D. $60$ min

Answer 1

Answer: (D)

$60$ min

$N_t=N_0\times {\left(\frac{1}{2}\right)}^n$

where, $N_t=$ after $n$ half-lives times volume concentration of $H_2{O}_2=1$

$N_0=$ initial concentration of $H_2{O}_2=16$ volume

$\therefore 1=16{\left(\frac{1}{2}\right)}^n$

$\frac{1}{16}={\left(\frac{1}{2}\right)}^4={\left(\frac{1}{2}\right)}^n$

$n=4$

Therefore, after $4$ half-lives times, volume of $H_2{O}_2$ solution becomes $1$ volume.

Time used on this procedure $=4\times 15min=60$ min.