A chemist whishes to prepare 100 liters of 45% purity of sulphuric acid .He has two kinds of acid solutions in stock ,one is 55% pure and the other is 30% pure .How many leters of each kind should be used for the mixture?

the chemist should use 60 liters of 55% solution and 40  litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

Step-by-step explanation:

From the given information,

Let x be the litres of 55% pure solution

Let y be the litres of 30% pure solution

Also;

Given that our total volume of solution is  100 litres

x+y =100   (1)

The total solution of pure by related by the sum of the individual pure concentrations to make up the concentration of final solution.

(0.55)(x)+(0.30)(y) = 0.45(100) (2)

From equation (1)

Let ; y = 100 - x

Replacing the value for y = 100 - x into equation (2)

(0.55)(x)+(0.30)(100-x) = 0.45(100)

0.55x + 30 - 0.30x = 45

0.55x - 0.30x = 45 - 30

0.25x = 15

x = 15/0.25

x = 60 liters of 55% solution

From ; y = 100 - x

y = 100 - 60

y = 40  litres of 30% solution.

Therefore, the chemist should use 60 liters of 55% solution and 40  litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

i believe

step-by-step explanation:

from my conclusion c

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