0 . 4 g of a metal was dissolved in 500 ml of N 10 H 2 SO 4 . When the reaction was over, the excess required was 150 ml N 15 alkali for complete neutralization. Calculate the equivalent mass of metal.

Given that 0 . 4 g of a metal was dissolved in 500 ml of N 10 H 2 SO 4 . When the reaction was over, the excess required was 150 ml N 15 alkali for complete neutralization. First, we should find out how much volume of H 2 SO 4 has been reacted. We know that, N 1 × V 1 = N 2 × V 2 0 . 1 × V 1 = 1 15 × 0 . 15 V 1 = 0 . 1 l = 100 ml remained The volume of acid reacted is 500 - 400 = 100 ml Number of moles of H 2 SO 4 reacted is given by the formula, moles of acid = molarity × volume of acid reacted moles of acid = N eq . number × 400 ml moles of acid = 0 . 1 n 1 × 0 . 4 l = 0 . 04 n 1 Now, moles of metal participated is given by the formula, moles of metal = given mass molar mass moles of metal = 0 . 4 molar mass moles of metal = 0 . 4 equivalent mass × n 2 But in a reaction, the product of a number of moles and the number of equivalents are the same. Hence, moles of acid × n 1 = moles of the metal × n 2 0 . 04 n 1 × n 1 = 0 . 4 equivalent mass × n 2 × n 2 Equivalent mass of metal = 0 . 4 0 . 04 = 10 g

0 . 12 gm of magnesium was dissolved in 500 ml of 0 . 1 N acid. What volume of N 2 NaOH solution will neutralise the resultant solution.

Given that 0 . 12 gm of magnesium was dissolved in 500 ml of 0 . 1 N acid. We know that, no . of moles × equivalent number = N × V Given mass molar mass × 2 = N × V 0 . 12 24 × 2 = N × V 0 . 01 = 0 . 1 × V V = 0 . 01 l = 100 ml Remaining volume is V 1 = 500 - 100 = 400 ml = 0 . 4 l We know that N 1 × V 1 = N 2 × V 2 0 . 1 × 0 . 4 = 0 . 5 × V 2 V 2 = 0 . 04 0 . 5 = 0 . 08 l = 0 . 08 × 1000 = 80 ml

A 0 . 50 g sample of impure CaCO 3 is dissolved in 50 ml of 0 . 985 N HCl . After the reaction is complete, the excess HCl requires 6 ml of 0 . 105 N caustic soda for neutralisation. Find the percentage of CaCO 3 in the sample.

Given that, 0 . 50 g sample of impure CaCO 3 is dissolved in 50 ml of 0 . 985 N HCl . After the reaction is complete, the excess HCl requires 6 ml of 0 . 105 N caustic soda for neutralisation. We know that, Equivalent weight = Normality × V ( i n m l ) Equivalent weight of H C l = equivalent weight of N a O H + Equivalent weight of C a C O 3 0.0985 × 5 0 = 0.105 × 6 + W C a C O 3 100 × 2 × 1000 4.925 - 0.630 = W CaCO 3 × 20 W CaCO 3 = 0.21475 g . Hence, equivalent weight of CaCO 3 = 0 . 21475 g . Percentage of CaCO 3 is, % CaCO 3 = 0.21475 0.50 × 100 = 42.95 %

Acid is generated in the stomach when the concentration of acid goes up, a man suffers from acidity. The doctor prescribes him antacid. By taking 3 antacid tablets, the strength decreases from N 10 to N 200 . If the fluid material is 100 ml , calculate the amount of base per antacid present per antacid tablet. (Equivalent mass of base in antacid = 100 ).

Normality is defined as the number of gram equivalents of solute , present in one litre of solution. Given that, normality of acid before reaction is N 1 = 1 10 . Volume of acid given is V = 100 ml = 0 . 1 l Let the number of moles of acid be m 1 and the equivalent weight of acid be n 1 . We know that, N 1 × V 1 = no . of moles × equivalent number 1 10 × 0 . 1 = m 1 × n 1 m 1 × n 1 = 0 . 01 m 1 = 1 100 × n 1 When the acid reacts with antacid, let the number of moles of acid be m 2 . Normality of acid after it reacts with antacid is N 2 = 1 200 . Volume of acid is 100 ml . We know that, N 2 × V 2 = no . of moles of acid after reaction × equivalent number 1 200 × 0 . 1 = m 2 × n 2 m 2 × n 2 = 0 . 002 m 2 = 1 2000 × n 2 The actual number of moles that take part in the reaction is given by, m = m 1 - m 2 m = 1 100 n - 1 2000 n m = 19 2000 n Let the equivalent number of base be n o . GIven number of tablets are 3 . We know that, No . of tablets × no . of moles of base per tablet × n o = eq . number of acid × no . of moles of acid reacted 3 × mass of base per tablet molar mass of base × n o = 19 2000 n × n 3 × mass of base per tablet molar mass of base n o = 19 2000 3 × mass of base per tablet equivalent mass of base = 19 2000 3 × mass of base per tablet = 19 2000 × 100 mass of base per tablet = 19 60 = 0 . 317