The volume, in mL , of 0.02 M K 2 Cr 2 O 7 solution required to react with 0.288 g of ferrous oxalate in acidic medium is (Molar mass of . Fe =56 g mol -1)

Question

The volume, in mL , of 0.02 M K 2 Cr 2 O 7 solution required to react with 0.288 g of ferrous oxalate in acidic medium is (Molar mass of . Fe =56 g mol -1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

undersetn =6 K2Cr2O7 + undersetn = 3FeC2O4 → Cr+3+ Fe+3 + CO2 (0.02 × 6 × V(mL)/1000) = (0.288/144) × 3 ⇒ V = 50 mL

Q. The volume, in $m L,$ of $0.02 M K_{2} C r_{2} O_{7}$ solution required to react with $0.288 g$ of ferrous oxalate in acidic medium is (Molar mass of $F e = 56 g m o l^{- 1})$

$n = 6 K_{2} C r_{2} O_{7} + n = 3 F e C_{2} O_{4} \to C r^{+ 3} + F e^{+ 3} + C O_{2}$

$\frac{0.02 \times 6 \times V ( m L )}{1000} = \frac{0.288}{144} \times 3$

$\Rightarrow V = 50 m L$

Q. The volume, in mL, of 0.02MK2​Cr2​O7​ solution required to react with 0.288g of ferrous oxalate in acidic medium is (Molar mass of Fe=56gmol−1)

n=6K2​Cr2​O7​​+n=3FeC2​O4​​→Cr+3+Fe+3+CO2​ 10000.02×6×V(mL)​=1440.288​×3 ⇒V=50mL

Q. The volume, in $m L,$ of $0.02 M K_{2} C r_{2} O_{7}$ solution required to react with $0.288 g$ of ferrous oxalate in acidic medium is (Molar mass of $F e = 56 g m o l^{- 1})$

$n = 6 K_{2} C r_{2} O_{7} + n = 3 F e C_{2} O_{4} \to C r^{+ 3} + F e^{+ 3} + C O_{2}$

$\frac{0.02 \times 6 \times V ( m L )}{1000} = \frac{0.288}{144} \times 3$

$\Rightarrow V = 50 m L$