Calculate the formula mass of$\mathit{C}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}$. What is the molarity of $\mathit{C}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}$in a solution containing 1.2gof $\mathit{C}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}$in a volume of 50 mL? How many grams of $\mathit{C}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}$are in 50 mLof $\mathbf{0}\mathbf{.}\mathbf{086}\mathit{M}\mathit{C}\mathit{a}\mathit{S}{\mathit{O}}_{\mathbf{4}}$?

Molarity$\mathit{=}\mathit{}\frac{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{v}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{L}}\mathit{}$

Atomic mass of$Ca=40.078amu$

Atomic mass of$S=32.065\mathrm{amu}$

Atomic mass of $O=15.999amu$

Formula mass of $CaS{O}_4$is calculated as,

Atomic mass of$Ca+$atomic mass of $S+4$(atomic mass of O)

$$\begin{gathered} =40.078amu+32.065amu+4(15.999amu) \\ =136.14amu \\ =136.14g/mol \end{gathered}$$

Sum of atomic mass of Calcium, Sulfur and four Oxygen atoms gives formula mass of Calcium sulfate$CaS{O}_4$

Hence, the formula mass of$CaS{O}_4$is$136.14g/mol$

Weight of$CaS{O}_4=1.2g$

Number of moles ofrole="math" localid="1663304367641" $CaS{O}_4=\frac{mass}{molarmass}$

$$\begin{gathered} =\frac{1.2g}{136.14g/mol} \\ =0.0088mol \end{gathered}$$

Volume of solution is$50\mathrm{mL}(=0.05\mathrm{L})$

Calculate the molarity of 50 mL solution containing 1.2 g of$CaS{O}_4$ as shown below,

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{CaSO}}_4}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}} \\ =\frac{0.0088\mathrm{mol}}{0.05\mathrm{L}} \\ =0.176\mathrm{M} \end{gathered}$$

Therefore, the molarity of solution containing 1.2 g of $CaS{O}_4$is 0.176M

Let ' x ' be the number of moles of$CaS{O}_4$ present in 50mL of$0.086MCaS{O}_4$ solution.

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{CaSO}}_4}{0.05\mathrm{L}} \\ 0.086M=\frac{x}{0.05\mathrm{L}} \\ x=0.05\mathrm{L}\times 0.086\mathrm{M} \\ \mathrm{x}=0.0043\mathrm{mol} \end{gathered}$$

Mass of$CaS{O}_4$ present in 50mL of$0.086MCaS{O}_4$ solution is calculated as shown

Number of moles of role="math" localid="1663304926877" $CaS{O}_4=\frac{mass}{molarmass}$

$0.0043mol=\frac{mass}{136.14/mol}$

Mass ofrole="math" localid="1663305000153" ${\mathrm{CaSO}}_4=0.0043\mathrm{mol}\times 136.14\mathrm{g}/\mathrm{mol}$

$=0.585g$

Therefore,50mL of$0.086MCaS{O}_4$ solution contains 0.585g of $CaS{O}_4$.