Calculate molarity of the resulting mixture, when 0.1 N of 100 ml $H_{2} S O_{4}$ is mixed with 0.2 N of 100 ml of NaOH solution:

25.0 M

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7.5 \times 10^{- 2} M

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15.0 \times 10^{- 2} M

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1.5 M

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Solution

The correct option is B $7.5 \times 10^{- 2} M$
$N = M \times n - f a c t o r$
$M_{H_{2} S O_{4}} = \frac{0.1}{2}, M_{N a O H} = 0.2$
$M_{3} (V_{1} + V_{2}) = M_{1} V_{1} - M_{2} V_{2}$
$M (100 + 100) = 0.2 \times 100 - \frac{0.1}{2} \times 100$
= 20 - 5
$= \frac{15}{200} M$
$= 7.5 \times 10^{- 2} M$

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