To neutralize completely 20 ml of 0.1 M phosphorus acid, 40 mol of KOH was required. What volume of this KOH solution will be required to neutralize 0.66 g of H_(3)PO_(2)?

To neutralize completely 20 ml of 0.1 M phosphorus acid, 40 mol of KOH

1.	To neutralize completely 20 ml of 0.1 M phosphorus acid, 40 mol of KOH was required. What volume of this KOH solution will be required to neutralize 0.66 g of H_(3)PO_(2)?
Answer» 100 ml 200 ml 300 ml 66.7 ml SOLUTION :Eqts of `H_(3)PO_(3)` = Eqts of KOH `20xx0.1xx2=1xx40xxMimpliesM=0.1` Now Eqts of KOH = Eqts of `H_(3)PO_(2)` `0.1xx1xxV=(0.66)/(66)impliesV=0.1" lit"` `V=100ml`