Question

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.

Answer 1

Answer: 0.20 M

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 1.40 M

$V_1$ = volume of stock solution = 72.0 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = 248 ml

$1.40\times 72.0=m\times 248$

$m=0.41M$

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

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.41 M

$V_1$ = volume of stock solution = 124 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = (124 +133) ml = 257 ml

$0.41\times 124=m\times 257$

$m=0.20M$

Thus the final concentration of the solution is 0.20 M.