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A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution

Respuesta :

Answer:

[tex]0.745~M[/tex]

Explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

Now, we have to identify the variables:

[tex]C_1~=~14.9~M[/tex]

[tex]V_1~=~25~mL[/tex]

[tex]C_2~=~?[/tex]

[tex]V_2~=~0.5~L[/tex]

Now, we have different units for the volume, so we have to do the conversion:

[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]

Now we can plug the values into the equation:

[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]

I hope it helps!