Answer to Question #141382 in General Chemistry for Mark

Question

A 57.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by adding 127 mL of water. What is the final concentration? Assume the volumes are additive.

Answer 1

$\begin{aligned} \textsf{molar concentration (C) }&= 1.50 \ M\\ \textsf{Volume (V) }&= 57.0 \ mL \\ &= 57 × 10^3 \ L\\ &= 5.7 × 10^{-2} \ dm^3\\ \\ \textsf{number of moles (n) } &= V × C\\ &= 5.7 × 10^{-2} × 1.50\\ &= 0.0855 \ moles \end{aligned}$

$\textsf{After first dilution, V }= 248mL = 0.248dm^3$

$\textsf{new molar concentration }= \dfrac{0.0855\ moles}{0.248\ dm^3} = 0.345M$

A 124mL portion is taken away

$\begin{aligned} V &= 124mL = 0.124\ dm^3\\ C &= 0.345M\\ \therefore \textsf{number of moles (n) }&= C × V\\ &= 0.345 × 0.124 \\ &= 0.0428\ moles \end{aligned}$

$\textsf{final volume } (V_{final}) = 124mL + 127mL = 251mL = 0.251dm^3$

$\begin{aligned} \therefore \textsf{final concentration } &= \dfrac{n}{V_{final}}\\ \\ &= \dfrac{0.0428\ moles}{0.251\ dm^3}\\ \\ &= 0.171M \end{aligned}$