What will be the normality of a solution obtained by mixing 0.45 N and 0.60 N NaOH in the ratio 1:2 by volume?

0.4N

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0.55N

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1.05N

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0.15N

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Solution

The correct option is B 0.55N
Normality of resulting solution $= \frac{N_{1} V_{1} + N_{2} V_{2}}{V_{1} + V_{2}}$
Since, $\frac{V_{1}}{V_{2}} = \frac{1}{2}$ ; $V_{2} = 2 V_{1}$ (substituting it in the above given equation)
$N_{1} = 0.45 N, V_{1} = V_{1}; N_{2} = 0.60 N, V_{2} = 2 V_{1}$ .
Now, $\frac{0.45 \times V_{1} + 0.60 \times 2 V_{1}}{V_{1} + 2 V_{1}}$
= $\frac{0.45 V_{1} + 1.20 V_{1}}{3 V_{1}}$
= $1.65 V_{1} / 3 V_{1}$ ( $V_{1}$ gets cancelled)
= 0.55 N
Option B is correct

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