1.26 g of crystalline oxalic acid H2C2O4. x H2O is dissolved to make 250 mL of solution. A 25.0 mL portion requires 20.4 mL of 0.0200 M KMnO4 in hot acid. Using molar mass of anhydrous H2C2O4 = 90 g mol^-1 and of H2O = 18 g mol^-1, find x to the nearest integer.

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JEE Advanced levelNumerical answerDifficulty 3 of 3eq074

$1.26 g$ of crystalline oxalic acid $H X_{2} C X_{2} O X_{4} \cdot x H X_{2} O$ is dissolved to make $250 mL$ of solution. A $25.0 mL$ portion requires $20.4 mL$ of $0.0200 M$ $KMnO X_{4}$ in hot acid. Using molar mass of anhydrous $H X_{2} C X_{2} O X_{4}$ = $90 g mo l^{- 1}$ and of $H X_{2} O$ = $18 g mo l^{- 1}$ , find $x$ to the nearest integer.

Numerical answer type.

Answer: 2

Oxalic acid in 25.0 mL = 2.5 x (0.0204 x 0.0200) = 1.020e-03 mol, so in 250 mL = 1.020e-02 mol. Molar mass of crystals = 1.26/1.020e-02 = 123.5 g/mol. x = (123.5 - 90)/18 = 1.86, i.e. 2.

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1.26 g of crystalline oxalic acid HX2​CX2​OX4​⋅xHX2​O is dissolved to make 250 mL of solution. A 25.0 mL portion requires 20.4 mL of 0.0200 M KMnOX4​ in hot acid. Using molar mass of anhydrous HX2​CX2​OX4​ = 90 gmol−1 and of HX2​O = 18 gmol−1, find x to the nearest integer.

Answer: 2

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