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$K_{s p}$ of $\mathrm{CaSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is $9 \times 10^{-6}$, find the volume for $1 \mathrm{~g}$ of $\mathrm{CaSO}_4$ (M.wt. = 136).
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Verified Answer
The correct answer is:
2.45 litre
$$
\begin{aligned}
& \text { } \mathrm{CaSO}_{4(s)} \rightleftharpoons \mathrm{Ca}^{2+}{ }_{(a q)}+\mathrm{SO}_4{ }^{2-}{ }_{(a q)} \\
& S \\
& K_{s p}=S^2=9 \times 10^{-6} \\
& S=3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$$
Solubility in $\mathrm{g}$ litre $^{-1}=$ molecular mass $\times S$
$$
=136 \times 3 \times 10^{-3}=408 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}
$$
$408 \times 10^{-3} \mathrm{~g}$ of $\mathrm{CaSO}_4$ present in 1 litre
$1 \mathrm{~g}$ of $\mathrm{CaSO}_4$ present is $\frac{1}{408 \times 10^{-3}}=2.45$ litre
\begin{aligned}
& \text { } \mathrm{CaSO}_{4(s)} \rightleftharpoons \mathrm{Ca}^{2+}{ }_{(a q)}+\mathrm{SO}_4{ }^{2-}{ }_{(a q)} \\
& S \\
& K_{s p}=S^2=9 \times 10^{-6} \\
& S=3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$$
Solubility in $\mathrm{g}$ litre $^{-1}=$ molecular mass $\times S$
$$
=136 \times 3 \times 10^{-3}=408 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}
$$
$408 \times 10^{-3} \mathrm{~g}$ of $\mathrm{CaSO}_4$ present in 1 litre
$1 \mathrm{~g}$ of $\mathrm{CaSO}_4$ present is $\frac{1}{408 \times 10^{-3}}=2.45$ litre
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