Answered

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

Sagot :

Answer

66.30 mL

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:

[tex]C_1V_1=C_2V_2[/tex]

Substitute the given parameters into the formula to get V₁

[tex]\begin{gathered} 5.724M\times V_1=2.063M\times183.95mL \\ \text{Divide both sides by 5.724M} \\ \frac{5.724M\times V_1}{5.724M}=\frac{2.063M\times183.95mL}{5.724M} \\ V_1=66.30\text{ mL} \end{gathered}[/tex]

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

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