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Q.

How much solid oxalic acid (Molecular weight 126) has to be weight to prepare 100 mL exactly 0.1 (N) oxalic solution in water?

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a

0.063 g

b

1.26 g

c

0.126 g

d

0.63 g

answer is C.

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Detailed Solution

Given: Molecular weight = 126g/mol

‘n’ factor = 2

$e q . w . t = \frac{126}{2} = 63 g / m o l .$

$M = \frac{W \times 1000}{m o l e c u l a r w e i g h t \times V o l}$

$0.1 = \frac{W \times 1000}{63 \times 100}$

$W = 0.63 g$

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