Question

What will be the normality of the salt solution obtained by neutralizing \(x \mathrm{ml}\) y \((\mathrm{N}) \mathrm{HCl}\) with y \(\mathrm{ml} \mathrm{x}(\mathrm{N}) \mathrm{NaOH}\), and fipally adding \((x+y)\) ml distilled water?
(A) \(\frac{2(x+y)}{x y} N\)
(B) \(\frac{x y}{2(x+y)} N\)
(C) \(\left(\frac{2 x y}{x+y}\right) \mathrm{N}\)
(D) \(\left(\frac{x+y}{x y}\right) N\)

Answer 1

Ans: (B)
$$
\begin{array}{lcccc}
& \multicolumn{2}{c}{\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}} \\
\text { number of m.e added } & \mathrm{xy} & \mathrm{xy} & 0 & 0 \\
\text { number of m.e left } & 0 & 0 & \mathrm{xy} & \mathrm{xy}
\end{array}
$$
Normality of salt solution \(=\frac{\text { number of milliequivalent }}{\text { Volume of solution in } \mathrm{mL}}=\frac{x y}{(x+y)+(x+y)} \mathrm{N}=\frac{x y}{2(x+y)} \mathrm{N}\)

Answer 2

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Answer 3

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