1g sample of H2O2 solution containing x% H2O2 by weight requires x mL of K2Cr2O7 solution forcomplete oxidation under acidic conditions. The molarity of K2Cr2O7 solution is —— .

3H2O2 + K2Cr2O2 + 4H2SO4 →K2SO4 + Cr2(SO4)3 + 7H2O + 3O2 M eq of H2O2 = m eq of K2Cr2O7100g of H2O2 solution contains x gof H2O2= Weightof H2O2m Eq of H2O2= Weightof H2O2×103EEwHH2O2=5.1x100×117×103=3xmEqCr2O72−≡H2O2mEq≡mEqN1V1(mL)≡3xN1×xmL=3xNK2Cr2O7=3NMK2Cr2O7=Nn=0.5M

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1g sample of H₂O₂ solution containing x% H₂O₂ by weight requires x mL of K₂Cr₂O₇solution for
complete oxidation under acidic conditions. The molarity of K₂Cr₂O₇ solution is —— .

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Detailed Solution

3H₂O₂ + K₂Cr₂O₂ + 4H₂SO₄ $\to$ K₂SO₄+ Cr₂(SO₄)₃ + 7H₂O + 3O₂
M eq of H₂O₂ = m eq of K₂Cr₂O₇

$\begin{aligned} 100 g of H_{2} O_{2} solution contains x gof H_{2} O_{2} \\ = Weightof H_{2} O_{2} \\ m Eq of H_{2} O_{2} = \frac{Weightof H_{2} O_{2} \times 10^{3}}{E {EwHH}_{2} O_{2}} \\ = \frac{5.1 x}{100} \times \frac{1}{17} \times 10^{3} = 3 x m E q \\ {Cr}_{2} O_{7}^{2 -} \equiv H_{2} O_{2} \\ m E q \equiv m E q \\ N_{1} V_{1} (m L) \equiv 3 x \\ N_{1} \times x m L = 3 x \\ N (K_{2} {Cr}_{2} O_{7}) = 3 N \\ M (K_{2} {Cr}_{2} O_{7}) = \frac{N}{n} = 0.5 M \end{aligned}$