Use Cr = 52. In alkaline solution H2O2 oxidises Cr^3+ to chromate: 2Cr^3+ + 3H2O2 + 10OH- -> 2CrO4^2- + 8H2O. Find the volume in mL (nearest integer) of 1.5 M H2O2 solution needed to oxidise 6.76 g of chromium present as Cr^3+.

JEE Advanced levelNumerical answerDifficulty 3 of 3rx157

Use Cr = 52. In alkaline solution $H X_{2} O X_{2}$ oxidises $Cr X^{3 +}$ to chromate: $2 Cr X^{3 +} + 3 H X_{2} O X_{2} + 10 OH X^{-} 2 CrO X_{4} X^{2 -} + 8 H X_{2} O$ . Find the volume in mL (nearest integer) of $1.5 M$ $H X_{2} O X_{2}$ solution needed to oxidise $6.76 g$ of chromium present as $Cr X^{3 +}$ .

Numerical answer type, answer in mL, tolerance 0.5.

n (Cr) = 6.76/52 = 0.13

mol. From the equation

Cr X^{3 +} : H X_{2} O X_{2} = 2 : 3

, so

H X_{2} O X_{2} = 0.13 \times 3/2 = 0.195

mol.

V = 0.195/1.5 = 130 mL