Question

A certain radioactive isotope ${}_Z^{A}X(t_{1/2}=100days)$ decays to ${}_Z^A{}_2^{8}Y$ . If 1 mole of ${}_Z^{A}X$ is kept in sealed container , how much he gas will accumulate at STP in 200 days?

• A. 16.8 litre
• B. 33.6 litre
• C. 22.4 litre
• D. 44.8 litre

Answer 1

The correct option is A 16.8 litre

$\because$ $K=\frac{1}{t}ln\frac{a_{\circ }}{a_t}\frac{l{n}_2}{t_{v_2}}=\frac{1}{t}ln\frac{a_{\circ }}{a_t}$

$\therefore \frac{l{n}_2}{100}=\frac{1}{100}ln\frac{a_{\circ }}{a_t}$

$⟹$ $ln\frac{a_{\circ }}{a_t}=2ln2⟹\frac{a_{\circ }}{a_t}=4$

$a_t=\frac{a_{\circ }}{4}=\frac{1}{4}$ mol undecayed,

gas accumulated $=(1-\frac{1}{4})\times 22.4=16.8L$