You have 58 g of calcium chloride (CaCl2) and you need to make a 5 M solution. You know that the molecular weight of Ca is 40 and that the molecular weight of Cl is 35.5. How many liters (L) of water would you need

Answers

Answer 1

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]


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The options that describe the experimental errors that would affect the percent of chloride reported are:

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Answer:

0.354 molal

Explanation:

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