Question

the mass of 70% pure H2SO4 Required for neutralisation of 1 mol of NaOH.....​

Answer 1

1 mole of H2SO4 requires 2 moles of NaOH for its complete neutralisation

mass of 100% pure H2SO4 required to neutralise 1 mole NaOH = 98/2=49g

thus mass of 70% pure H2SO4=x

49/x × 100=70

x=70g

Ans-(c) 70g