Chemistry - Chemical Kinetics Question with Solution | TestHub

${ }_{79}{\mathrm{A}\mathrm{u}}^{198}\stackrel{\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }-B\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }}{\to }{ }_{80}{\mathrm{H}\mathrm{g}}^{198}$
$\text{k}\text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\text{=}\frac{\text{0}\text{.693}}{\text{65}}$
After 260 hr,
$\text{k}\text{=}\frac{\text{2}\text{.303}}{\text{260}}\text{log}\frac{\text{a}}{\text{a}-\text{x}}$
$\frac{\text{0}\text{.693}}{\text{65}}\text{=}\frac{\text{2}\text{.303}}{\text{260}}\text{log}\frac{\text{a}}{\text{a}-\text{x}}$
$\frac{\text{a}}{\text{a}-\text{x}}\text{=}\text{16}$
$\frac{\text{1}}{\text{1}-\text{x}}\text{=}\text{16}$
$\text{x}\text{=}\frac{\text{15}}{\text{16}}\text{g}\text{=}\text{0}\text{.9375 g}$