Đáp án:

$\rm \%^{35}Cl=100\%-24,23\%=75,77\%$

$\rm M_{Cl}=\dfrac{37×24,23+35×75,77}{100}=35,4846\ (amu)$

Cho $\rm M_{Al}=27\ (amu)$.

$\rm M_{AlCl_3}=133,4538\ (amu)$

Giả sử có 1 mol $\rm AlCl_3$.

Bảo toàn nguyên tố Cl:

$\rm n_{Cl}=3n_{AlCl_3}=3\ (mol)$

$\to\begin{cases}\rm n_{^{37}Cl}=0,7269\ (mol)\\\rm n_{^{35}Cl}=2,2731\ (mol)\end{cases}$

$\rm\%m_{^{37}Cl}=\dfrac{0,7269×37}{133,4538}×100\%=20,15\%\\\%m_{^{35}Cl}=\dfrac{2,2731×35}{133,4538}×100\%=59,62\%$