Answer
$0.92mg$
Work Step by Step
We can find the mass of the isotope as follows:
$\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}$
$\lambda=\frac{\ln 2}{2.70 d}=0.256d^{-1}$
We know that
$N=\frac{R}{\lambda}$
$\implies N=\frac{225 Ci}{0.2567d^{-1}}(\frac{86,400s}{d})(\frac{3.7\times 10^{10}s^{-1}}{1Ci})=2.8\times 10{18}$
Now we can find the mass as follows:
$m=M\frac{N}{N_A}$
We plug in the known values to obtain:
$m=(198g)\frac{2.80\times 10^{18}}{6.022\times 10^{23}}=0.92mg$
