Question

500 ml of 0.1 M KCl, 200 ml of 0.01 M $NaN{O}_3$ and 500 ml of 0.1 M $AgN{O}_3$ was mixed. What is the total number of moles of ions in solution?

• A. 204 mmol
• B. 104 mmol
• C. 54 mmol
• D. 72 mmol

Answer 1

The correct option is B

104 mmol

In this question, $AgN{O}_3$ will react with KCl and AgCl forms a precipitate.
We know that,
Molarity = $\frac{Numberofmolesofsolute}{Numberoflitresofsolution}$
No. of moles of KCl $=0.1\times \frac{500}{1000}=0.05$
No. of moles of $NaN{O}_3=\frac{200}{1000}\times 0.1=0.002$

No. of moles of nitrate ion from $AgN{O}_3=N{O}_3^{-}=\frac{500}{1000}\times 0.01=0.05$
Hence, number of moles of AgCl precipitated = 0.05

Total nitrate ions: 0.05 + 0.002 = 0.052
Hence, in solution, only K+ ions and $N{O}_3^{-}$ will be present apart from $N{a}^{+}$.

Total moles = 0.05 + 0.052 + 0.002 = 0.104 or 104 mmoles