HNO₃ used as a reagent in stoichiometric analysis has specific gravity of 1.5 and contains 63% HNO₃ by mass. The volume of magnesium hydroxide solution which is 2.9 g/L in Mg(OH)₂ required for neutralizing 20 mL of above HNO₃ solution is?

2HNO₃+Mg(OH)₂=Mg(NO₃)₂+2H₂O.

Mass of HNO₃ taken is 1.5*20/1000 gm = 0.03 gm.

Actual mass of HNO₃  present 63% of 0.03 gm = 0.0189 gm = 0.0189/63 moles = 0.0003 mole.

Moles of H⁺ present = 0.0003 .

Moles of OH^- needed (from Mg(OH)₂) = 0.0003.

So, Moles of Mg(OH)₂ needed = 0.00015 = 0.008745 gm Mg(OH)₂ .

Let volume needed be v mL.

Then, 2.9*v/1000 = 0.008745

=> v = 3.01.

Ans:3.01 g of 100% pure Mg(OH)₂ solution will be needed.