Question

It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and 88 wt.% water. Two solutions are available, the first contains 5 wt.% ethanol, and the second contains 25 wt.% ethanol. How much of each solution are mixed to prepare the desired solution?

Answer 1

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

Step-by-step explanation:

Let the mass of the first solution be x and second solution be y.

Amount solution required = 1250 kg

x + y = 1250 kg....[1]

Percentage of ethanol in required solution = 12% of 1250 kg

Percentage of ethanol in solution-1 = 5% of x

Percentage of ethanol in required solution = 25% of y

5% of x + 25% of y =12% of 1250 kg

`(5)/(100)* x+(25)/(100)y=(12)/(100)* 1250 kg`

x + 5y = 3000 kg...[2]

Solving [1] and [2] we :

x = 437.5 kg , y = 812.5 kg

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.