Question

A chemistry teacher has a flask of a 45% acidic solution and another bottle of 28% acidic solution.

How much would he need to use of each container to arrive at a final solution that is 100 mL of a 33.95% solution?

Answer 1

35 ml of 28% acidic solution and 65ml of 45% acidic solution is to be used

Explanation:

Let the number of ml of the 45% solution be x while that of the 28% solution be y

Mathematically;

x + y = 100 ••••••(i)

Furthermore

45% of x + 28% of y = 33.95% of 100

0.45x + 0.28y = 33.95 ••••••(ii)

From i,

x = 100-y

put this into ii

0.45(100-y) + 0.28y = 33.95

45 - 0.45y + 0.28y = 33.95

45 - 33.95 = 0.45y - 0.28y

11.05 = 0.17y

y = 11.05/0.17

y = 65

Recall

x = 100 - y

x = 100 - 65

x = 35 ml