CBSE Class 11-science Answered
The mass of 70% Sulphuric acid by mass required to neutralize 1 mole of NaOH
Asked by | 21 Mar, 2013, 10:35: AM
Expert Answer
2NaOH + H2SO4 = Na2SO4 + H2O
Pure H2SO4 required for 1 mole of NaOH = 1/2 mole = 49g
70% H2SO4 required for 1 mole = 49 x 100 /70
= 70 g
Answered by | 03 May, 2013, 07:35: PM
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