Question

0.2 mole of HCl and 0.1 mole of $\text{CaC}{\text{l}}_{\text{2}}$ were dissolved in water to have 500 mL of solution, the molarity of $\text{C}{\text{l}}^{-}$ ions is:

• A. 0.04 M
• B. 0.8 M
• C. 0.4 M
• D. 0.08 M

Answer 1

The correct option is B 0.8 M

Number of moles of HCl is 0.2 mole.

$\text{HCl}\to {\text{H}}^{+}+\text{C}{\text{l}}^{-}$

Thus, 0.2 mole of HCl gives 0.2 moles of chloride ions.

Number of moles of $\text{CaC}{\text{l}}_2$ is 0.1 mole

$\text{CaC}{\text{l}}_2\to \text{C}{\text{a}}^{2+}+2\text{C}{\text{l}}^{-}$

Thus, 0.1 mole of $\text{CaC}{\text{l}}_2$ gives $\left(0.1\times 2\right)$ mole of chloride ions.

0.1 mole of $\text{CaC}{\text{l}}_2$ gives 0.2 mole of chloride ions.

Therefore, total number of moles for $\text{C}{\text{l}}^{-}$ ions is $\left(0.2+0.2\right)$ mole that is 0.4 mole

Now, molarity $=\frac{\text{Number of moles}}{\text{Volume}\left(L\right)}$

Volume gives is 500 mL (0.5L)

Molality = $\frac{0.4}{0.5}$

$\text{Molality}=0.8\text{M}$

Hence, the correct option is $B$