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5.3 grams of Na2CO3 is dissolved in 200ml of a solution. It is diluted by adding 800 ml of water. 100ml of the resulting solution requires ‘V’ ml of 0.02N H2SO4 solution for complete neutralization. ‘V’ is

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answer is 3.

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Detailed Solution

i) N = 5.353×1000200 N = 0.5Nii) V1N1 = V2N2 2000.5 = 200+800N2 N2 = 0.1Niii) VbNb = VaNa 1000.1 = Va0.02 Va = 500ml

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