Using K=39, Cr=52, O=16 (molar mass of K2Cr2O7=294), the mass of K2Cr2O7 needed in acidic medium to oxidise 0.0240 mol of Fe^2+ to Fe^3+ is:

JEE Main levelSingle correct optionDifficulty 2 of 3eq190

Using $K = 39$ , $Cr = 52$ , $O = 16$ (molar mass of $K X_{2} Cr X_{2} O X_{7} = 294$ ), the mass of $K X_{2} Cr X_{2} O X_{7}$ needed in acidic medium to oxidise $0.0240$ mol of $Fe X^{2 +}$ to $Fe X^{3 +}$ is:

Each

Cr X_{2} O X_{7} X^{2 -}

takes 6 electrons and each

Fe X^{2 +}

gives 1, so

1

mol

K X_{2} Cr X_{2} O X_{7}

oxidises 6 mol

Fe X^{2 +}

. Moles needed

= 0.0240/6 = 0.00400

; mass

= 0.00400 \times 294 = 1.176 g