(Q25).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 100 ml of a 68% solution?

Let the first solution contains 50% acid.

Second solution contains 80% acid.

Let x ml of 1st solution and y ml of second solution are added.

Then x + y =

Acid content in the ‘mix’ is 50% of x + 80% of y = 68%

x
100
 +
y
100
 = ⇒
x
 +
y
 =
x+y
10
 = ⇒ x + y = ................(2)
equation (1) × ⇒ x + y =
equation (2)         ⇒ x + y =
-y = -
∴ y =
=

Substituting y = in equation (1) we get

x + = 100

⇒ x = 100 – =

∴ Quantity of first solution = ml

Quantity of second solution = ml