Question

What volume of 6M HCl and 2M HCl should be mixed to get two litres of 3M HCl?

Answer 1

Suppose the volume of 6M HCl required to prepare 1 L of 3M HCl = x L

Volume of 2M HCl required = 1 - x L

Apply molarity equation

M1V1 + M2V2 = M3V3

6 � (x) + 2 ( 1-x) = 3 x 1

6x + 2 - 2x = 3

4x +2 = 3

4x = 3 - 2

4x = 1

x = 1/4 = 0.25 L

so, volume of 6M HCl required = 0.25 L

volume of 2M HCl required = 1 - 0.25 = 0.75 L

Answer 2

Given, 6M HCL mixed with 2M HCL to get 2L of 3M HCL.

To find: How much volume.

We know that, ${ M_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }={ M }_{ 3 }{ V }_{ 3 }$

$6\quad \times \quad x\quad +\quad 2(2-x)\quad =\quad 3\times 2\\6x\quad +\quad 4\quad -\quad 2x\quad =\quad 6\\ 4x\quad =\quad 2\\ x=0.5L$

$(2-x)=2-0.5L=1.5L$

0.5L of 6M Hydrochloric acid (HCL) is mixed with 1.5L of 2M Hydrochloric acid (HCL) to get 2L of 3M Hydrochloric acid (HCL).