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100 ml of solution containing 0.1 mole of NaOH per litre was mixed with 100 mL solution containing 0.02 mole of $H_{2} S O_{4}$ per litre. The amount of NaOH in the mixture in grams will be

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0.36

0.24

2.4

0.12

answer is B.

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Detailed Solution

2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O

Milliequivalent of $N a O H = 0.1 \times 100 \times 1 = 10$

Milliequivalent of $H_{2} S O_{4} = 0.02 \times 100 \times 2 = 4$

After neutralization remaining millimoles of $N a O H = 10 + 4 = 6$

$M a s s = 6 \times 10.3 \times 40 = 0.24 g$

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