Question

100 g of a sample of common salt containing contamination of $N{H}_{4}Cl$ (Molar mass = 53.5 g/mol) and $MgC{l}_2$ (Molar mass = 95 g/mol) to the extent of 2% each by mass is dissolved in water. Which of the following is/are correct when volume of 5% by mass of $AgN{O}_3$ (Molar mass = 170 g/mol) solution (d = 1.04 ${\text{g/cm}}^3)$ is required to precipitate all chloride ions?

• A. Molarity of $AgN{O}_3$ solution = 0.3 M
• B. Volume of $AgN{O}_3$ required to precipiate all chloride ions = 570 L
• C. Molarity of $AgN{O}_3$ solution = 0.03 M
• D. Volume of $AgN{O}_3$ required to precipiate all chloride ions = 5.7 L

Answer 1

Answer: (A), (D)

The correct options are
A Molarity of $AgN{O}_3$ solution = 0.3 M
D Volume of $AgN{O}_3$ required to precipiate all chloride ions = 5.7 L

$\text{Amount of}N{H}_{4}Cl=2\text{g}=\frac{2}{53.5}=0.037\text{mol}$
$C{l}^{-}\text{ions in}N{H}_{4}Cl=0.037\text{mol}$
$\text{Amount of}MgC{l}_2=2\text{g}=\frac{2}{95}=0.021\text{mol}$
$C{l}^{-}\text{ions in}MgC{l}_2=2\times 0.021=0.042\text{mol}$
$\text{Amount of}NaCl=96\text{g}=\frac{96}{58.5}=1.63\text{mol}$
$\text{Total}C{l}^{-}\text{ions}=0.037+0.042+1.63=1.71\text{moles}$

$AgN{O}_3+C{l}^{-}\to AgCl+N{O}_3^{-}$
From the reaction we can see that ,
$\text{Moles of}AgN{O}_3\text{required for precipitation}=1.71\text{mol}$
$\text{Molarity of}AgN{O}_3=\frac{\text{\%}\times d\times 10}{\text{Molar mass}}=\frac{5\times 1.04\times 10}{170}=0.3\text{M}$
$\therefore \text{Volume of}AgN{O}_3=\frac{1.71}{0.3}=5.7\text{L}$