## Chemistry and Chemical Reactivity (9th Edition)

- No, the compound does not completely dissolve. Only $2.1 \times 10^{-4}$ $g$ of it will dissolve
1. Calculate the molar mass $(RaSO_4)$: 226* 1 + 32.07* 1 + 16* 4 ) = 322.07g/mol 2. Calculate the number of moles $(RaSO_4)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.025}{ 322.07}$ $n(moles) = 7.8\times 10^{- 5}$ 3. Find the concentration in mol/L $(RaSO_4)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 7.8\times 10^{- 5}}{ 0.1}$ $C(mol/L) = 7.8\times 10^{- 4}$ ---------- Now, calculate the molar solubility: 1. Write the $K_{sp}$ expression: $RaSO_4(s) \lt -- \gt 1Ra^{2+}(aq) + 1S{O_4}^{2-}(aq)$ $4.2 \times 10^{-11} = [Ra^{2+}]^ 1[S{O_4}^{2-}]^ 1$ 2. Considering a pure solution: $[Ra^{2+}] = 1S$ and $[S{O_4}^{2-}] = 1S$ $4.2 \times 10^{-11}= ( 1S)^ 1 \times ( 1S)^ 1$ $4.2 \times 10^{-11} = 1S^ 2$ $4.2 \times 10^{-11} = S^ 2$ $\sqrt [ 2] {4.2 \times 10^{-11}} = S$ $6.5 \times 10^{-6} = S$ - This is the molar solubility value for this salt. Since the concentration we added is greater than the solubility value for this salt, there will be precipitation. The molar solubility is how much of it will dissolve. Now, we have to convert the molar solubility into grams: 3. Find the number of moles: $Concentration(M) = \frac{n(mol)}{V(L)}$ $6.5 \times 10^{-6} = \frac{n(mol)}{0.1}$ $6.5 \times 10^{-6} * 0.1 = n(mol)$ $6.5 \times 10^{-7} moles = n(mol)$ 4. Calculate the mass $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $322.07 * 6.5 \times 10^{-7} = mass(g)$ $2.1 \times 10^{-4} = mass(g)$