A solution of 8% hydrochloric acid is to be diluted by adding a 2% hydrochloric acid solution to it. The resulting mixture is to be more than 4% and less than 6% of hydrochloric acid. If we have 640 litres of the 8% solution and Q denotes the quantity of 2% solution that will be added, then find the value of Q2-Q1/320 where Q1 and Q2, marks the minimum and maximum value of Q.
It is of linear inequalities and not of chemistry......
Answers
Answer:
Q2-Q1/320 = 3
Step-by-step explanation:
A solution of 8% hydrochloric acid is to be diluted by adding a 2% hydrochloric acid solution to it. The resulting mixture is to be more than 4% and less than 6% of hydrochloric acid.
640 Litre of 8% hydrochloric acid solution
Q1 min value of 2% hydrochloric acid solution
Resultant = 6 %
in 640 Litre hydrochloric acid = (8/100) * 640 = 51.2 Litre
in Q1 litre hydrochloric acid = (2/100)Q1 = 0.02Q1 Lire
(51.2 + 0.02Q1)/(640 + Q1) = 6/100
=> 5120 + 2Q1 = 3840 + 6Q1
=> 1280 = 4Q1
=> Q1 = 320
Q2 maz value of 2% hydrochloric acid solution
Resultant = 4 %
in 640 Litre hydrochloric acid = (8/100) * 640 = 51.2 Litre
in Q2 litre hydrochloric acid = (2/100)Q1 = 0.02Q2 Lire
(51.2 + 0.02Q2)/(640 + Q2) = 4/100
=> 5120 + 2Q2 = 2560 + 4Q2
=> 2560 = 2Q2
=> Q2 = 1280
Q2-Q1/320 = (1280 - 320)/320 = 4 - 1 = 3
Q2-Q1/320 = 3
Answer:
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