Question

Equal volumes of $0.1\text{M}AgN{O}_3$ and $0.2\text{M}NaCl$ are mixed. The concentration of $N{O}_3^{-}$ ions in the mixture will be:

• A. $0.1\text{M}$
• B. $0.05\text{M}$
• C. $0.20\text{M}$
• D. $0.15\text{M}$

Answer 1

The correct option is B $0.05\text{M}$

Let the volume of both, $AgN{O}_3$ and $NaCl$ be $V\text{L}$.
$\text{Molarity}=\frac{\text{number of moles}}{\text{volume of the solution in L}}$
Number of moles of $AgN{O}_3=0.1\text{V}$
Moles of $A{g}^{+}=\text{moles of}N{O}_3^{-}=0.1\text{V}$
So, molarity of $N{O}_3^{-}=\frac{\text{total moles of}N{O}_3^{-}}{\text{total volume of the solution in L}}=\frac{0.1\text{V}}{2\text{V}}=0.05\text{M}$