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

An exess of `NaOH` was added to `100 mL` of a `FeCl_(3)` solution which gives `2.14 "of" Fe (OH)_(3)`. Calculate the normality of `FeCl_(3)` solution.

A

`0.2 N`

B

`0.3 N`

C

`0.6 N`

D

`1.8 N`

Text Solution

Verified by Experts

The correct Answer is:
C
`underset({:("excess"):})(3NaOH) + underset({:(100 mL),(1 "mmol" -= 1 "mmol"):})(FeCl_(3)) rarr Fe(OH)_(3) + 3NaCl`
`FeCl_(3) -= Fe (OH)_(3)`
`M xx 100 -= (2.14)/(107) xx 1000`
`M_(FeCl_(3)) = 0.2`
`N_(FeCl_(3)) = 0.2 xx 3` (Valency factor = 3)
`= 0.6 N`
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