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

Question

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

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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`

4 views Answered 2023-02-05 15:29