Question

In an aqueous solution of barium nitrate, the $\left[N{O}_3^{-}\right]$ is 0.080 M. This solution can be labeled as

• A. $0.040N\left(Ba\right(N{O}_2{)}_2$
• B. $0.160MBa(N{O}_3{)}_2$
• C. $0.080NBa(N{O}_3{)}_2$
• D. $0.080MN{O}_3^{-}$

Answer 1

The correct option is C $0.080NBa(N{O}_3{)}_2$

In$Ba(N{O}_3{)}_{2+}$ the molar ratio of $Ba(N{O}_3{)}_2$ to $N{O}_3^{-}$ is 1 : 2
Therefore, the molarity of the $Ba(N{O}_3{)}_2$ solution is
$\frac{1}{2}\times 0.080=0.040M$
As n - factor of $Ba(N{O}_3{)}_2$ is 2, its normality will be 0.080 N as normality= molarity $\times$ valency factor.
So, the correct answer will be option (c).