A 100mL sample of water was treated to convert any iron present to ${Fe}^{2 +}$ 25mL of $0.002 {M K}_{2} {Cr}_{2} O_{7}$
resulted in the reaction.
$6 {Fe}^{2 +} + {Cr}_{2} O_{7}^{2 -} + 14 H^{+} ⟶ 6 {Fe}^{3 +} + 2 {Cr}^{3 +} + 7 H_{2} O$
The excess $K_{2} {Cr}_{2} O_{7}$ was back-titrated with $7.5 mL$ of $0.01 M {Fe}^{2}$ solution. Calculate the parts per million (ppm) of iron in the water sample.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 126.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

milli equivalent of excess $K_{2} {Cr}_{2} O_{7} = 7.5 \times 0.01 \times 1$

milli equivalent of $K_{2} {Cr}_{2} O_{7}$ taken $= 25 \times 0.002 \times 6$

milli equivalent of ${Fe}^{2 +}$ in water sample

$\begin{array}{l} = 25 \times 0.002 \times 6 - 7.5 \times 0.01 \times 1 \\ = 0.225 \end{array}$

$∴$ Mass of ${Fe}^{2 +} = 0.225 \times 10^{- 3} \times 56 = 12.6 \times 10^{- 3} g$

$100 g ⟶ 12.6 \times 10^{- 3}$

$10^{6} g ⟶ \frac{12.6 \times 10^{- 8}}{100} \times 10^{6} = 126$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE