Respuesta :
Answer:
0.1045 L (4 s.f.)
Explanation:
The molecular weight, [tex]M_{r}[/tex], of calcium chloride is
[tex]M_{r}(\mathrm{CaCl_{2}}) \ = \ A_{r}(\mathrm{Ca}) \ + \ 2 \times A_{r}(\mathrm{Cl}) \ = \ (40 \ + \ 2 \times 35.5) \ \mathrm{g \ mol^{-1}} \ = \ 111 \ \mathrm{g \ mol^{-1}}[/tex], where [tex]A_{r}[/tex] is the atomic weight of the respective elements.
To calculate the number of moles (n) of calcium chloride present in 58g,
[tex]n_{\mathrm{CaCl_{2}}} \ = \ mass \ \div \ molecular \ weight \ (M_{r}) \\ \\ \-\hspace{1.13cm} = \ 58g \ \div \ 111 \ \mathrm{g \ mol^{-1}} \\ \\ \-\hspace{1.13cm} = \ 0.5225 \ \mathrm{mol \ of \ CaCl_{2}}[/tex]
Since molarity, M, is defined as the number of moles (n) of the solute per liters of a solution (L), hence to get the volume (V) required to make up a 5 molar solution is
[tex]n \ = \ M \times V \\ \\ V \-\hspace{0.05cm} = \ \displaystyle\frac{n}{M} \\ \\ V \-\hspace{0.05cm} = \displaystyle \frac{0.5225 \ \mathrm{mol}}{5 \ \mathrm{mol \ L^{-1}}} \\ \\ V \-\hspace{0.05cm} = \ 0.1045 \ \mathrm{L}[/tex]
