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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.
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