Question

14) How many mL of 5.724 M HNO3 are needed to prepare a 183.95 mL of 2.063 M HNO3?

Answer 1

66.30 mL

Step-by-step explanation

Given parameters:

Initial concentration, C₁ = 5.724 M

Final volume, V₂ = 183.95 mL

Final volume, V₂ = 183.95 mLFinal concentration, C₂ = 2.063 M

What to find:

The initial volume, V₁

Step-by-step solution:

Using the dilution law:

`C_1V_1=C_2V_2`

Substitute the given parameters into the formula to get V₁

`\begin{gathered} 5.724M* V_1=2.063M*183.95mL \\ \text{Divide both sides by 5.724M} \\ (5.724M* V_1)/(5.724M)=(2.063M*183.95mL)/(5.724M) \\ V_1=66.30\text{ mL} \end{gathered}`

Hence, 66.30 mL of 5.724 M HNO3 are needed to prepare a 183.95 mL of 2.063 M HNO3?