A known mass of oxalic acid dihydrate is dissolved in water to make 100 ml solution. 10 ml of this solution is diluted to 100 ml. 20 mL of the solution requires 20 mL of 0.1 N NaOH for complete neutralisation. (eq. mass of oxalic acid dihydrate = 63). The mass of oxalic acid in gram in the original solution is

Q. 20 ml of 0.02 M

K M n O_{4}

was required to completely oxidise 10 ml of oxalic acid solution. What is the molarity of the oxalic acid solution?

Q. 100 mL of a 0.1 M HCl solution, 200 mL of a 0.2 M HCl solution, 300 mL of a 0.3 M HCl solution, 400 mL of a 0.4 M HCl solution and 500 mL of a 0.5 M HCl solution are mixed. 200 mL of this solution is diluted with 300 mL of water. What is the molarity of HCl in the resultant solution?

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500 ml of a solution contains 12.6 gm of oxalic acid. 10 ml of this solution is diluted to 100 ml in a flask. What is the molarity of the solution?

The correct option is C 0.02Molarity of oxalic acid,C2H2O4.2H2O=12.61260×1000500 =15 M M1=0.2V1=10mlV2=100mlSince the number of equivalents should be the same.N1V1=N2V2M1×2×V1=M2×2×V20.2×10=M2×100M2=0.210M2=0.02M