Question

A 52.0 mL aliquot of a 1.20M solution is diluted to a total volume of 208mL. A 104 mL portion of that solution is diluted by adding 139mL of water. what is the final concentration? assume the volumes are additive.

Answer 1

As number of moles of solute remains constant throughout the dilution;

Moles in 52 ml solution$=52 \times 1.2 \times 10^{-3}=52 \times 1.2\space mili mol$ which is equal to the number of moles in 208 ml solution.

Moles in 104 solution$=\frac{1}{2}\times 52 \times 1.2 \times 10^{-3} moles$ $=26 \times 1.2 \times 10^{-3} moles$

Final volume $=104+139=243 ml$

Final concentration$=\frac{ 26 \times 1.2 \times 10^{-3}}{243 \times 10^{-3}}=0.128M$