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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.
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