Chemistry - Chemical Kinetics Question with Solution | TestHub

${\mathrm{t}}_{1/2}=138.4$days, t = 69.2 day

Number of half-lifes$\mathrm{n}=\frac{\mathrm{t}}{{\mathrm{t}}_{1/2}}=\frac{69.2}{138.4}=\frac{1}{2}$

Amount of Po left$\frac{1}{2}$after half-life$=\frac{1}{{\left(2\right)}^{1/2}}\mathrm{g}=0.707\mathrm{ }\mathrm{g}$

Amount of Po used in$\frac{1}{2}$half-life$=1-0.707=0.293\mathrm{ }\mathrm{g}$

Now${}_{84}\mathrm{P}{\mathrm{o}}^{210}\to {}_{82}\mathrm{P}{\mathrm{b}}^{206}+{}_2{\mathrm{H}\mathrm{e}}^4$

210 g Po on decay will produce = 4g He

0.293 g Po on decay will produce$=\frac{4\mathrm{ }\times \mathrm{ }0.293\mathrm{ }}{210}=5.581\times {10}^{-3}\mathrm{g}\mathrm{ }\mathrm{H}\mathrm{e}$

Volume of He at STP$=\frac{5.581\mathrm{ }\times \mathrm{ }{10}^{-3}\mathrm{ }\times \mathrm{ }22400}{4}=31.25\mathrm{ }\mathrm{m}\mathrm{L}=31.25\mathrm{ }\mathrm{c}{\mathrm{m}}^3$