Question

3. The amount of wet NaOH containing 20% water
required to neutralise 6 litre of 0.5 M H2SO4
solution is​

Answer 1

Wet Na OH having 15% water or 85 purity.

Molar mass of Na OH is 40g/mol.

70 liters of 0.5 N solution needs 70*0.5

                                                     = 35 moles

                                                     = 35 *40

                                                     = 1,400 g

Hence, the amount of wet NaOH is 1400 * 100/85

                                                       =16.470 * 100

                                                        =1647 g of wet Na OH