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25 cm^(3) of oxalic acid completely neut...

`25 cm^(3)` of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

A

0.064

B

0.045

C

0.015

D

0.032

Text Solution

Verified by Experts

The correct Answer is:
D
`undersetoverset(|)(COOH)(C )OOH + 2NaOH to undersetoverset(|)(COONa)(C ) OONa + 2H_2O`
Moles of 0.064g of NaOH
` = (0.064)/(40) = 0.0016` mol
2 moles of NaOH react with 1 mol of oxalic acid
` therefore ` 0.0016 mol of NaOH react with oxalic acid
` = (0.0016)/(2) = 8 xx 10^(-4) ` mol
volume of solution = 25 mL
Molarity of oxalic acid solution ` = (8 xx 10^(-4) xx 1000)/(25)`
0.032 M
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