A chemistry laboratory requests for a 25% solution of ferrous sulphate. A supplier has 40 millilitres of 20% solution. How many millilitres of 40% solution should be added to make it a 25% solution (correct to two decimal places)?
- A16.4
- B15.2
- C13.33
- D14.3
Solution & Step-by-step Explanation
Let x be the volume (in ml) of the 40% solution to be added.
The total amount of pure ferrous sulphate before and after mixing must remain equal:
Pure substance in 20% solution+Pure substance in 40% solution=Pure substance in final 25% solution
20% of 40+40% of x=25% of (40+x)
0.20×40+0.40×x=0.25×(40+x)
8+0.40x=10+0.25x
0.40x−0.25x=10−8
0.15x=2
x=
0.15
2
=
15
200
=
3
40
≈13.33ml
The total amount of pure ferrous sulphate before and after mixing must remain equal:
Pure substance in 20% solution+Pure substance in 40% solution=Pure substance in final 25% solution
20% of 40+40% of x=25% of (40+x)
0.20×40+0.40×x=0.25×(40+x)
8+0.40x=10+0.25x
0.40x−0.25x=10−8
0.15x=2
x=
0.15
2
=
15
200
=
3
40
≈13.33ml