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.

Options

A. `0.2 N`
B. `0.3 N`
C. `0.6 N`
D. `1.8 N`

Answer

C

Explanation

`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`