A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by adding 133 mL of water. What is the final concentration? Assume the volumes are additive.

Question

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Respuesta :

kobenhavn kobenhavn · Answer 1 · 2019-09-08T15:04:32+02:00

Answer: 0.20 M

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock solution = 1.40 M

[tex]V_1[/tex] = volume of stock solution = 72.0 ml

[tex]M_2[/tex] = molarity of diluted solution = m

[tex]V_2[/tex] = volume of diluted solution = 248 ml

[tex]1.40\times 72.0=m\times 248[/tex]

[tex]m=0.41M[/tex]

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock solution = 0.41 M

[tex]V_1[/tex] = volume of stock solution = 124 ml

[tex]M_2[/tex] = molarity of diluted solution = m

[tex]V_2[/tex] = volume of diluted solution = (124 +133) ml = 257 ml

[tex]0.41\times 124=m\times 257[/tex]

[tex]m=0.20M[/tex]

Thus the final concentration of the solution is 0.20 M.