# How many liters of a 3M solution have to be mixed with 2 liters of a 4M solution to yield a 3.6M solution?

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We have a 3M solution and a 4M solution and we are supposed to mix them to create a 3.6M solution. Now if 2 liters of the 4M solution are used, let the volume of the 3M solution used be equal to L.

The total volume of the final solution is 2 + L. As we need a 3.6M solution, there should be 3.6*(2 + L) moles in the solutions we use. The number of moles available in the constituent solutions is 4*2 + 3*L

We now equate 4*2 + 3*L to 3.6*(2 + L)

=> 4*2 + 3*L = 3.6*(2 + L)

=> 8 + 3L = 7.2 + 3.6L

=> 8 – 7.2 = 3.6L – 3L

=> .8 = .6L

=> L = .8 / .6

=> L = 4/3 = 1.33 liters

**Therefore the volume of the 3M solution required is 4/3 liters.**

Let us cosider the solution contents of 4M solution of 2 litres = 4*2= 8

Let x be the litres of 3M solution to be mixed with 2 litre of 4M solution. The x litre of solution contents = 3x.

Then the totaotal volume of the mixed solution = 2+x litres.

The contents of the solution of x+2 litres of mixed 3.6 M solution = (x+2)*3.6M.

Therefore the required equation is:

(x+2)3.6 = 2*4 + 3x.

3.6x +7.2 = 8 +3x

3.6x-3x = 8-7.2

0.6x = 0.8

x = 0.8/0.6 = 4/3 litre.

Therefore 4/3 litres of 3M solution should be mixed with 2 lires of 4M solution to get mexed solution of 3.6M.