A vessel contains 20 % concentrated solution of Sulphuric acid. If 50 ml more sulphuric acid is added in it, then concentration of Sulphuric acid in vessel becomes 36 % What is the volume of sulphuric acid in original mixture in vessel?
Answers
Answer:
Volume of Sulphuric acid in original mixture = 40 ml
Step-by-step explanation:
Vessel contains 20% Concentrated Solution of Sulphuric acid
Let say total solution = 100 S ml
Sulphuric acid = (20/100)100S = 20S ml
50 ml more sulphuric acid added
now Sulphuric acid = 20S + 50 ml
Total Solution = 100S + 50 ml
now sulphuric acid = 36%
(36/100)(100S + 50) = 20S + 50
36S + 18 = 20S + 50
16S = 32
S = 2
Slphuric acid = 20S = 20*2 = 40 ml
Volume of Sulphuric acid in original mixture = 40 ml
Answer:
Step-by-step explanation:
Let volume of solution in the vessel be 'y' ml
Volume of sulphuric acid in this solution = 20 % of y ml
= 0.2 y ml
Volume of water in the solution = y - 0.2 y = 0.8 y
Now, we are adding 50 ml of sulphuric acid into the solution (because of which the concentration of the solution will change) however the amount of water will remain same as before.
Volume of sulphuric acid = (50 + 0.2y) ml
Volume of water = 0.8y ml
Volume of solution = 50 + 0.2y + 0.8y = (50 + y) ml
Strength of solution = 36% = (Volume of sulphuric acid/Volume of solution) x 100
⇒0.36 = (50 + 0.2y)/(50 + y)
⇒ 0.36(50 + y) = 50 + 0.2 y
⇒18 + 0.36 y = 50 + 0.2 y
⇒ 0.16 y = 38
Volume of solution originally present (y) = 237.5 ml
Volume of sulphuric acid = 20% of 237.5 = 0.2 x 237.5 = 47.5 ml