Question

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
​

Answer 1

Step-by-step explanation:

Let the first solution contains 50% acid Let second solution contains 80% acid Let the first solution be x ml and second solution y ml added x + y = 100 ..... (1)

The acid content in the mixture of y 50% of x + 80% of y

100

50x

+

100

80y

=68⇒

2

x

+

5

4y

=68 ⇒

10

5x+8y

=68⇒5x+8y=680...(2) (1)×5⇒5x+5y=500....(3) (2)⇒5x+8y=680....(4) (3)−(4)gives −3y=180⇒y=

−3

−180

=60 Substituteyvaluein(1) ∴ The quantity of first solution = 40 ml ∴ The quantity of second solution = 60 ml

Answer 2

Answer:

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