(Sửa đề: $30,905g$ axit boric $H_3BO_3$)
Gọi $\%^{10}{\rm{B}}=x(\%)\Rightarrow \%^{11}{\rm{B}}=100-x(\%)$
$\overline{A}=10,81$
$\Rightarrow 10.x\%+11(100-x)\%=10,81$
$\Leftrightarrow x=19\%$
$\Rightarrow \%^{11}{\rm{B}}=100-19=81\%$
$n_{H_3BO_3}=\dfrac{30,905}{3+10,81+16.3}=0,5(mol)=n_B$
$\Rightarrow n_{^{11}{\rm{B}}}=0,5.81\%=0,405(mol)$
$\to m_{^{11}{\rm{B}}}=0,405.11=4,455g$