Question

A chemist has two solutions hydrochloric acid in stock. One is 50%solution and the other is 80% . How much of each should be used to be obtain 100ml of a 68% solution

Answer 1

let x ml taken from 50% solution and y ml taken from 80 % solution.
'y'ml = (100 - x) ml [since total sol is 100 ml]
now 50 % of x + 80% of (100 - x) = 68 % of 100
1/2 x + 4/5 × 100 - 4/5× x = 68/100 × 100
0.5 x + 80 - 0.8x = 68
-0.3 x = 68- 80= -12
x = 12/ 0.3
x = 60 ml.
then y = 40 ml.

Answer 2

Step-by-step explanation:

let x ml taken from 50% solution and y ml taken from 80 % solution.

'y'ml = (100 - x) ml [since total sol is 100 ml]

now 50 % of x + 80% of (100 - x) = 68 % of 100

1/2 x + 4/5 × 100 - 4/5× x = 68/100 × 100

0.5 x + 80 - 0.8x = 68

-0.3 x = 68- 80= -12

x = 12/ 0.3

x = 60 ml.

then y = 40 ml.

HOPE ITS HELPFUL : )