\(0.1 M \quad\quad 0.2 M \quad0.1 M\)
\(BaSO _4 \rightleftharpoons Ba ^{+2}+ SO _4{ }^{2-}\)
\(a - S \quad S \quad S +0.1 \approx 0.1\)
\(K _{ SP }= S \times 10^{-1}\)
\(\Rightarrow 1 \times 10^{-10}= S \times 10^{-1}\)
\(\Rightarrow S =10^{-9} mol L^{-1 }\)
So, \(S =10^{-9} \times 233\,g\,L ^{-1}\)
So, Answer : \(233\)
$NH_3$ + $H_2O$ $\rightleftharpoons$ $NH_4^ + + O{H^ - }$