We can determine the volume of solution by equation the number of equivalents of the solution. 1 1 2 2 N V =N V Let 3 1 1 3 2 2 N =0.5 N V = 100 cm N = 0.1 N V = x cm Then the volume is determined as, 3 1 1 2 2 N V 0.5 100 V 500 cm N 0.1 x The final volume of the diluted solution is 3 500 cm . If 3 100 cm the volume is provided by the sulphuric acid then the volume of water to be added is, 3 volume of water = 500 -100 = 400 cm Thus, (a) is the correct option.
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