The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of NaOH is: (Molar Mass of S = 32 gmol−1, O = 16 gmol−1 and Na = 23 gmol−1)
A
100 g
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B
98 g
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C
200 g
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D
196 g
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Solution
The correct option is A 100 g H2SO4+2NaOH→Na2SO4+H2O
According to the stoichiometry, 2 moles of NaOH required 1 mole of sulphuric acid. Mass of sulphuric acid required = 98 g Mass of 98% sulphuric acid ×98100=98 ∴ Mass of 98 % pure sulphuric acid required for the titration = 100 g