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What volume of 1.75 M hydrochloric acid (HCl aq) must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid?

Sagot :

Sebassandinidn

Answer:

[tex]V_1=0.0714L[/tex]

Explanation:

Hello there!

In this case, since we need to dilute the 1.75-M HCl, and the total number of moles remain unchanged, we can write:

[tex]n_1=n_2[/tex]

And in terms of volume and concentration:

[tex]C_1V_1=C_2V_2[/tex]

Thus, we can solve for the volume of the concentrated HCl as shown below:

[tex]V_1=\frac{C_2V_2}{C_1}[/tex]

Therefore, we plug in the data to get:

[tex]V_1=\frac{0.250M*0.500L}{1.75 M}\\\\V_1=0.0714L[/tex]

Best regards!

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