2.56 × 10 - 3 equivalent of KOH is required to neutralise 0 .12544 g H 2 XO 4 . The atomic mass of X (in g/mol ) is: [Given: H 2 XO 4 is a dibasic acid]

Question

2.56 × 10 - 3 equivalent of KOH is required to neutralise 0 .12544 g H 2 XO 4 . The atomic mass of X (in g/mol ) is: [Given: H 2 XO 4 is a dibasic acid], 16, 8, 7, 32

MARKS App · Accepted Answer

Let the atomic weight of X = y No. of equivalents of KOH = No. of equivalents of H 2 XO 4 i.e. 0 .12544 66 + y 2 = 0 .12544 ∴ y = 32

Search any question & find its solution

Looking for more such questions to practice?