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An excess of NaOH was added to 100 mL of...

An excess of NaOH was added to 100 mL of a ferric chloride solution.This caused the precipitation of 1.425 g of `Fe(OH)_3`.Calculate the normality of the ferric chloride solution.

A

0.20 N

B

0.50 N

C

0.25 N

D

0.40 N

Text Solution

Verified by Experts

The correct Answer is:
D
`3NaOH+FeCl_3toFe(OH)_3+3NaCl`
mili eq. of NaOH = mili eq. of `Fe(OH)_3`
So, for `Fe(OH)_3 " " 100xxN=W/Exx1000(E_(Fe(OH)_3)=("mol. Wt.")/3)`
`N=(1.425xx10xx3)/107=0.399=0.4 N`
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