Answer
Molar solubility: $1.13 \times 10^{-18}M$
Mass in 15 L : $1.62 \times 10^{-15} g$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ CuS(s) \lt -- \gt 1Cu^{2+}(aq) + 1S^{2-}(aq)$
$1.27 \times 10^{-36} = [Cu^{2+}]^ 1[S^{2-}]^ 1$
2. Considering a pure solution: $[Cu^{2+}] = 1S$ and $[S^{2-}] = 1S$
$1.27 \times 10^{-36}= ( 1S)^ 1 \times ( 1S)^ 1$
$1.27 \times 10^{-36} = S^ 2$
$ \sqrt [ 2] {1.27 \times 10^{-36}} = S$
$1.13 \times 10^{-18} = S$
- This is the molar solubility value for this salt.
3. Find the number of moles:
$Concentration(M) = \frac{n(mol)}{V(L)}$
$1.13 \times 10^{-18} = \frac{n(mol)}{15}$
$1.13 \times 10^{-18} * 15 = n(mol)$
$1.695 \times 10^{-17} moles = n(mol)$
4. Determine the molar mass of this compound (CuS):
63.55* 1 + 32.07* 1 = 95.62g/mol
5. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 95.62 * 1.695 \times 10^{-17} = mass(g)$
$1.62 \times 10^{-15} = mass(g)$
