A student in a chemistry laboratory has access to two acid solutions. The first one is 20% acid and the second solution is 45% acid.(The percentages are by volume). How many millilitres of each solution should the student mix together to obtain 100ml of a 30% acid solution?

For the preparation of the 30% 100 ml acid solution, 40 ml 45% solution has been mixed with 60 ml 20% acid solution.

To prepare the solution with the concentrated solution:

MV = M1V1 + M2V2

The M has been the concentration of the resulted solution, and V has been the volume of the resulting solution.

M1 and M2 and V1 and V2 have been the concentration and volume of the given solution.

0.30 × 100 ml = 0.20 × x + 0.45 × (100 - x)

30 = 0.20x + 45 - 0.45x

0.45x - 0.20x = 45 - 30

0.25x = 15

x = 60 ml

The volume of 20% solution has been x = 60 ml

The volume of 45% solution has been 100 - x = 40 ml.

For the preparation of the 30% 100 ml acid solution, 40ml 45% solution has been mixed with 60ml 20% acid solution.

For more information about the acid preparation, refer to the link:

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Volume of each solution : 60 ml 20% and 40 ml 45%

Given

20% and 45% acid

100 ml of 30% acid

Required

Volume of each solution

Solution

Molarity from 2 solutions :

Vm Mm = V₁. M₁ + V₂. M₂  

m = mixed solution

V = volume

M = molarity

V₁ = x ml

V₂ = (100 - x) ml

Input the value :

100 . 0.3 = x . 0.2 + (100-x) . 0.45

30 = 0.2x+45-0.45x

0.25x=15

x= 60 ml

V₁ = 60 ml

V₂ = 100 - 60 = 40 ml

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