Question

200 mL of a 0.25 M $NaCl$ solution, 250 mL of a 0.5 M $CaC{l}_2$ solution and 100 mL of a 0.60 M $AlC{l}_3$ solution are mixed together and diluted to a final volume of 750 mL. Determine the molarity of chloride ions in solution assuming all the three salts are completely dissociated.

• A. 0.38 M
• B. 0.78 M
• C. 0.64 M
• D. 1.24 M

Answer 1

The correct option is C 0.64 M

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in mL}}\times 1000$

$\text{Moles of solute}=\frac{\text{(molarity (M) × volume in mL)}}{1000}$

$\text{Moles of}NaCl=\frac{0.25\times 200}{1000}=0.05$

Since, $NaCl$ has $1C{l}^{-}$ ions
Moles of $C{l}^{-}$ ions = 0.05

$\text{Moles of}CaC{l}_2=\frac{0.5\times 250}{1000}=0.125$

Since, $CaC{l}_2$ has $2C{l}^{-}$ ions
Moles of $C{l}^{-}$ ions = 0.25
$\text{Moles of}AlC{l}_3=\frac{0.60\times 100}{1000}=0.06$

Since, $AlC{l}_3$ has $3C{l}^{-}$ ions
Moles of $C{l}^{-}$ ions = 0.180
$\text{Total moles of}C{l}^{-}\text{ions}=0.05+0.25+0.180$
$=0.48\text{mol}$
$\text{Molarity of}C{l}^{-}\text{ions}=\frac{0.48}{750}\times 1000=0.64M$