Question
The solubility product of $$AgCl$$ is $$1.5625 \times {10^{ - 10}}$$ at $$25{\,^ \circ }C.$$ Its solubility in grams per litre will be
A.
$$143.5$$
B.
$$108$$
C.
$$1.57 \times {10^{ - 8}}$$
D.
$$1.79 \times {10^{ - 3}}$$
Answer :
$$1.79 \times {10^{ - 3}}$$
Solution :
$$AgCl \rightleftharpoons \mathop {A{g^ + }}\limits_s + \mathop {C{l^ - }}\limits_s $$
$${s^2} = 1.5625 \times {10^{ - 10}};$$ $$s = 1.25 \times {10^{ - 5}}\,mol\,{L^{ - 1}}$$
$$\eqalign{
& {\text{Solubility in}}\,\,g\,{L^{ - 1}} \cr
& = {\text{Molar mass}}\, \times \,s \cr
& = 143.5 \times 1.25 \times {10^{ - 5}} \cr
& = 1.79 \times {10^{ - 3}}\,g\,{L^{ - 1}} \cr} $$