A volume of 30 ml of a solution containing 9.15 g per litre of an oxalate K x H y ( C 2 O 4 ) z ⋅ n H 2 O is required for titrating 27 ml of 0.12 N - NaOH and 36 ml of 0.12 N - KMnO₄ separ

Question

A volume of 30 ml of a solution containing 9.15 g per litre of an oxalate K x ​ H y ​ ( C 2 ​ O 4 ​ ) z ​ ⋅ n H 2 ​ O is required for titrating 27 ml of 0.12 N - NaOH and 36 ml of 0.12 N - KMnO₄ separately. Assume all H-atoms are replaceable and x, y and z are in the simple ratio of g-atoms. The value of xyzn is

TestHub · Accepted Answer

Mass in 30 ml = 9.15×0.03 = 0.2745 g. Acid eq = 27×0.12 = 3.24 meq → acid eq wt = 0.2745/0.00324 = 84.72; redox eq = 36×0.12 = 4.32 meq → redox eq wt = 63.54. Ratio 2z/y = 84.72/63.54 = 4/3, so y:z = 3:2; charge balance x+y=2z gives x=1, i.e. KH₃(C₂O₄)₂. Molar mass = 3×84.72 = 254.2 = 218+18n, so n=2. Hence xyzn = 1322.

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