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How much concentrated 15.54 M sulfuric acid is need to prepare 6.1 mL of a 2.28 M solution?

User Shkarik
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1 Answer

4 votes

Answer:

0.895 mL

Step-by-step explanation:

Using the formula;

C1V1 = C2V2

Where;

C1 = initial concentration (M)

C2 = final concentration (M)

V1 = initial volume (mL)

V2 = final volume (mL)

According to the information provided in this question;

C1 = 15.54M

V1 = ?

C2 = 2.28M

V2 = 6.1 mL

Using C1V1 = C2V2

V1 = C2V2 ÷ C1

V1 = (2.28 × 6.1) ÷ 15.54

V1 = 13.908 ÷ 15.54

V1 = 0.895 mL

User Soham
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