$10$ ml of $N H_{4} C l$ , $20$ ml of $N / 2 H_{2} S O_{4}$ and $30$ ml of $N / 3 H N O_{3}$ are mixed together and volume made to one litre. Normality of resulting solution is:

30 N / 100

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N / 10

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N / 20

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N / 40

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Solution

The correct option is A $30 N / 100$
As we know $M (e q) = N \times V$

where
M(eq) = equivalent molarity
N = normality of solution
V = volume of solution

Applying this formula for each we get

For $N H_{4} C l - M (e q) = 1 \times 10 = 10 M$

For $H_{2} S O_{4} - M (e q) = 1 / 2 \times 20 = 10 M$

For $H N O_{3} - M (e q) = 1 / 3 \times 30 = 10$ M

For total soluion, $M (e q) = M_{1} + M_{2} + M_{3}$

$M (e q) = 10 + 10 + 10 = 30 M$

and volume = 1000 mL

Applying same formula $M (e q) = N \times V$

$30 = N \times 1000$

$N = 30 / 100$

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