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Q.

The weight of NaOH required to neutralize 20ml of 0.05M oxalic acid is

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a

0.2g

b

0.8g

c

0.04g

d

0.08g

answer is D.

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Detailed Solution

$N_{1} V_{1} = N_{2} V_{2} \frac{w}{g m w} x \frac{1000}{V 1} x V_{1} = 0.05 x 20 x 2 \frac{w}{40} x 1000 = 0.05 x 20 x 2 w = 0.08 g$

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The weight of NaOH required to neutralize 20ml of 0.05M oxalic acid is