A volume of 30 ml of a solution containing 7.4 g per litre of an oxalate N a x H y ( C 2 O 4 ) z ⋅ n H 2 O is required for titrating 15 ml of 0.1 N - NaOH and 30 ml of 0.1 N - KMn O 4 se

Question

A volume of 30 ml of a solution containing 7.4 g per litre of an oxalate N a x ​ H y ​ ( C 2 ​ O 4 ​ ) z ​ ⋅ n H 2 ​ O is required for titrating 15 ml of 0.1 N - NaOH and 30 ml of 0.1 N - KMn O 4 ​ 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 (Na = 23)

TestHub · Accepted Answer

Mass in 30 ml = 7.4 × 0.03 = 0.222 g. Acid eq = 15 × 0.1 = 1.5 meq → acid eq wt = 0.222/0.0015 = 148; redox eq = 30 × 0.1 = 3 meq → redox eq wt = 74. Ratio 2z/y = 148/74 = 2, so y = z ; charge balance x+y = 2z gives x = y = z = 1, i.e. NaH C 2 ​ O 4 ​ . Molar mass = 148 = 112 + 18n, so n = 2. Hence xyzn = 1112.

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