A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100ml of a 68% solution.
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Let the proportion of the first solution be represented by x and the second solution be represented by y.
first solution contains 50% solution = 50% of x
= 50 × x/100 = x/2
second solution contains 80% solution = 80% of y
= 80 × y/100 = 4y/5
given, together they can give 100ml of 68% solution
e.g., x/2 + 4y/5 = 68% of 100
x/2 + 4y/5 = 68
5x + 8y = 680 .......(i)
also, total percentage of solution concentration is 100% .
then, x + y = 100 .......(ii)
solve equations (i) and (ii),
multiplying 5 with equation (ii) and subtracting from (i),
5x + 8y - 5x - 5y = 680 - 500
3y = 180 => y = 60ml
and x = 40 ml
hence, 1st solution = 40ml and second solution = 60ml
first solution contains 50% solution = 50% of x
= 50 × x/100 = x/2
second solution contains 80% solution = 80% of y
= 80 × y/100 = 4y/5
given, together they can give 100ml of 68% solution
e.g., x/2 + 4y/5 = 68% of 100
x/2 + 4y/5 = 68
5x + 8y = 680 .......(i)
also, total percentage of solution concentration is 100% .
then, x + y = 100 .......(ii)
solve equations (i) and (ii),
multiplying 5 with equation (ii) and subtracting from (i),
5x + 8y - 5x - 5y = 680 - 500
3y = 180 => y = 60ml
and x = 40 ml
hence, 1st solution = 40ml and second solution = 60ml
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