Question

1) if you have 8.00M sulfuric acid solution, and you took 20ml of this stock solution and diluted to 500mL total, what is the concentration of the dilute solution? discuss your steps when you give the answer

2) If you have 8.00M of sulfuric acid solution and you want to dilute to 400mL diluted solution with concentration of 2.0M; what volume of the stock solution you should use? what volume of solvent you need to add when you dilute it

Answer 1

In both questions applying the condition of

Number of moles before dilution = number of moles after dilution

n₁ = n₂

we know that

Concentration = number of moles/volume of solution

C = n/V

n = CV

So we have

C_1V1 = C₂V₂

Question (1)

C1 = concentration of sulphuric acid before dilution = 8.00M

V1 = volume of solution before dilution = 20ml

C2= concentration of sulphuric acid after dilution

V2 = volume of solution after dilution = 500ml

Putting the all value in formula

(8.00M)(20ml) = (C2)(500ml)

(C2) = 0.32M

So concentration of sulphuric acid after dilution is 0.32M

Question 2

C1 = concentration of sulphuric acid before dilution = 8.00M

V1 = volume of solution before dilution

C2 = concentration of sulphuric acid after dilution = 2.00M

V2 = volume of solution before dilution = 400ml

Putting the all value in formula

(8.00M)(V2) = (2.00M)(400ml)

(V2) = 100ml

so volume of 8.00M sulphuric acid is 100ml