Question

Find pH of 3g of NaOH dissolved in 400 ml of water

Answer 1

Full ScreenSolved ExampleEx.1 The molarity of 20% (W/W) solution of sulphuric acid is 2.55 M. The density of the solution is :(A) 1.25 g cm-3(B) 0.125 g L-1(C) 2.55g cm-3(D) unpredictable(Ans. A)Sol.Volume of 100 g of solution =mlM =or d == 1.249 » 1.25Ex.2 The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Calculate the molarity and normality of the solution-(A) 1.445 M (B) 14.45 M (C) 144.5 M (D) 0.1445 M(Ans. A)Sol.Volume of 100 gram of the solution ==mL =litre=litreNumber of moles of H2SO4in 100 gram of the solution =Molarity ==×= 1.445 MEx.3 Calculate the molarity of pure water (d = 1g/L)(A) 555 M (B) 5.55 M (C) 55.5 M (D) None(Ans. C)Sol.Consider 1000 mL of waterMass of 1000 mL of water= 1000 × 1 = 1000 gramNumber of moles of water == 55.5Molarity === 55.5 MEx.4 Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml of0.1 M solution-(A) 2.65 gram (B) 4.95 gram (C) 6.25 gram (D) None(Ans. A)Sol.We know thatMolarity =where;W = Mass of Na2CO3in gramM = Molecular mass of Na2CO3in grams =106V =Volume of solution in litres == 0.25Molarity =Hence , ==or W == 2.65 gramEx.5 The freezing point of a 0.08 molalaqueous solution of NaHSO4is _ 0.372°C. The dissociation constant forthe reactionHSO4-HSO42_is(A) 4 × 10-2(B) 8 × 10-2(C) 2 × 10-2(D)1.86 × 10-2(Ans. A)Sol.DTf= Kf× mm =This means that total molal conc. of all particles is 0.2.NaHSO4¾→ NaHSO4-