Question

The molarity of $C{l}^{-}$ ion in a 100 mL aqueous solution which has 2 % $\left(\frac{w}{v}\right)NaCl$, 4 % $\left(\frac{w}{v}\right)CaC{l}_2$ and 6 % $\left(\frac{w}{v}\right)N{H}_{4}Cl$ will be:

• A. 0.342 M
• B. 0.721 M
• C. 1.121 M
• D. 2.181 M

Answer 1

The correct option is D 2.181 M

100 mL solution contains 2 g $NaCl$, 4 g $CaC{l}_2$ and 6 g $N{H}_{4}Cl$
Moles of $NaCl=\frac{2}{58.5}=0.034$
$\therefore \text{Moles of}C{l}^{-}\text{from}NaCl=0.034\text{mol}.......\left(i\right)$
$\text{Moles of}CaC{l}_2=\frac{4}{111}=0.036\text{mol}$
$\therefore \text{Moles of}C{l}^{-}\text{from}CaC{l}_2=2\times 0.036=0.072\text{mol}.......\left(ii\right)$
$\text{Moles of}N{H}_{4}Cl=\frac{6}{53.5}=0.112\text{mol}$
$\therefore \text{Moles of}C{l}^{-}\text{from}N{H}_{4}Cl=0.112\text{mol}.......\left(iii\right)$
$\text{Molarity}=\frac{\text{Moles of}C{l}^{-}}{\text{Volume in L}}=\frac{0.034+0.072+0.112}{0.1}$

$\text{Molarity}=2.18\text{M}$