Excess of NaOH (aq) was added to 100 mL of `FeCI_(3)` (aq) resulting into 2.14 g of `Fe(OH)_(3).` The molarity of `FeCI_(3)` (aq) is:
(Given molar mass of Fe = 56g `mol^(-1)` and molar mass of CI = 35.5g `mol^(-1)`)

Options

A. 0.2 M
B. 1.8 M
C. 0.3 M
D. 0.6 M

Answer

A

Explanation

`{:(3NaOH+,FeCl_(2),rarr,Fe(OH)_(3),+,3NaCl,because,"Moles of " Fe(OH)_(3) = (2.14)/(107) = 2 xx 10^(-2)),(,"100 mL",,"2.14 g",,,,):}`
`:.` Moles of `FeCl_(3)` = moles of `Fe(OH)_(3) = 2 xx 10^(-2)`, Now, `M = ("no. of moles " xx 1000)/("volume (mL)") = (2 xx 10^(-2))/(100) xx 1000`
=0.2 M