Answer :
Suppose that the volume of 6M HCl (V1) = x L
So, Volume of 2M HCl required to make 1L of 3M HCl (V2) = (1-x) L
Let us take M1 = 6M and M2 = 2M,
We know that, Molarity of a mixture equation :
(M1V1 + M2V2) / (V1 + V2) = M3
[6x + 2(1-x)] / 1 = 3
6x - 2x + 2 = 3
4x = 1 x = 1/4 = 0.25 L
Therefore, Volume of 6M HCl = 0.25L and Volume of 2 M HCl = 1-0.25 = 0.75 L
So, Volume of 2M HCl required to make 1L of 3M HCl (V2) = (1-x) L
Let us take M1 = 6M and M2 = 2M,
We know that, Molarity of a mixture equation :
(M1V1 + M2V2) / (V1 + V2) = M3
[6x + 2(1-x)] / 1 = 3
6x - 2x + 2 = 3
4x = 1 x = 1/4 = 0.25 L
Therefore, Volume of 6M HCl = 0.25L and Volume of 2 M HCl = 1-0.25 = 0.75 L