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The amount of oxalic acid (mol. wt. 63) ...

The amount of oxalic acid (mol. wt. 63) required to prepare 500 mL of its 0.10 N solution is

A

0.315 g

B

3.150 g

C

6.300 g

D

63.00 g

Text Solution

Verified by Experts

The correct Answer is:
B
1000 mL of 1 N oxalic acid sol = 63 g
500 mL of 0.1 N oxalic acid sol `=63/1000 xx 500 xx 0.1 = 3.15 g`
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