JEE MAIN - Chemistry (2020 - 5th September Evening Slot - No. 1)
The volume, in mL, of 0.02 M K2Cr2O7 solution
required to react with 0.288 g of ferrous
oxalate in acidic medium is _______.
(Molar mass of Fe = 56 g mol–1)
(Molar mass of Fe = 56 g mol–1)
Answer
50
Explanation
K2Cr2O7 + FeC2O4 $$ \to $$ Cr+3 + Fe+3 + CO2
nfactor of K2Cr2O7 = 3 $$ \times $$ 2 = 6
nfactor of FeC2O4 = 1 + 2 = 3
m. eq. of K2Cr2O7 = m. eq. of FeC2O4
$$ \Rightarrow $$ $${{6 \times 0.02 \times vol} \over {1000}}$$ = $${{0.288} \over {144}} \times 3$$
$$ \Rightarrow $$ vol = 50 ml
nfactor of K2Cr2O7 = 3 $$ \times $$ 2 = 6
nfactor of FeC2O4 = 1 + 2 = 3
m. eq. of K2Cr2O7 = m. eq. of FeC2O4
$$ \Rightarrow $$ $${{6 \times 0.02 \times vol} \over {1000}}$$ = $${{0.288} \over {144}} \times 3$$
$$ \Rightarrow $$ vol = 50 ml
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