Question

For a mixture of $100$mL of $0.3$ M $CaC{l}_2$ solution and $400$ mL of $0.1$M $HCl$ solution, select the correct options.

• A. Total concentration of cations $=0.14$M
• B. $\left[C{l}^{-}\right]=0.2$M
• C. Moles of $C{a}^{2+}$ in mixture $=0.03$
• D. None of these

Answer 1

The correct option is C Moles of $C{a}^{2+}$ in mixture $=0.03$

$1$ mole $CaC{l}_2$ and $1$ mole $HCl$ give $1$ mole $C{a}^{2+}$ and $1$ mole $H^{+}$ respectively.

So total concentration of cations $(C{a}^{2+}$ and $H^{+})$

$=\frac{0.3M\times 100ml+0.1M\times 400ml}{100ml+400ml}=0.14M$

$\Rightarrow \left[C{l}^{-}\right]=\frac{2\times 0.3\times 100+0.1\times 400}{100+400}=0.2M$

Moles of $C{a}^{2+}$ in mixture= $\frac{0.3\times 100}{1000}=0.03$ moles