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