Question

What volume of 12 N hcl and 3N hcl must be mixed to get 6N hcl?

Answer 1

Suppose that the volume of 12 HCl (V1) = x L

So, Volume of 3N HCl required to make 1L of 6N HCl (V2) = (1-x) L

Let us take M1 = 12N and M2 = 3N,

We know that, Molarity of a mixture equation :

(N1V1 + N2V2) / (V1 + V2) = N3

[12x + 3(1-x)] / 1 = 6

12x - 3x + 3 = 6

9x = 3 = 1/3= 0.33 L

Therefore, Volume of 12N HCl = 0.33L and Volume of 3N HCl = 1-0.33 = 0.67 L

Answer 2

Explanation:

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