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A 54.0 mL aliquot of a 1.60 M solution is diluted to a total volume of 218 mL. A 109 mL portion of that solution is diluted by adding 161 mL of water. What is the final concentration? Assume the volumes are additive g

Respuesta :

Answer: The final concentration is 0.16 M.

Explanation:

According to the dilution law:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock solution = 1.60 M

[tex]V_1[/tex] = volume of stock  solution = 54.0 ml

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

[tex]V_2[/tex] = volume of diluted solution = 218 ml

Putting these values:

[tex]1.60\times 54.0=M_2\times 218[/tex]

[tex]M_2=0.40M[/tex]

Now 109 ml of this diluted solution is further diluted by adding 161 mL of water.

Again applying dilution law:

[tex]0.40\times 109=M_3\times (109+161)[/tex]

[tex]0.40\times 109=M_3\times 270[/tex]

[tex]M_3=0.16M[/tex]

Thus the final concentration is 0.16 M

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