How do I calculate the molarity of a solution prepared by diluting 87.44 mL of 0.743 M potassium chloride to 150.00 mL?

Jun 30, 2017

$\text{Concentration"="Moles of solute"/"Volume of solution}$

Here, $\text{concentration} \cong 0.4 \cdot m o l \cdot {L}^{-} 1$

Explanation:

And we have initially, $87.44 \times {10}^{-} 3 \cdot L \times 0.743 \cdot m o l \cdot {L}^{-} 1 = 0.0650 \cdot m o l$, and we divide this molar quantity by the NEW volume to get the NEW concentration....

(87.44xx10^-3*Lxx0.743*mol*L^-1)/(150*mLxx10^-3*L*mL^-1)=??*mol*L^-1.

Alternatively, we could use the old expression.......

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, i.e.

${C}_{2} = \frac{{C}_{1} {V}_{1}}{{V}_{2}} = \frac{0.743 \cdot m o l \cdot {L}^{- 1} \times 87.44 \cdot \cancel{m L}}{150.00 \cdot \cancel{m L}}$

Which of course is the same quotient.........which explicity gives an answer with units of $m o l \cdot {L}^{-} 1$, as required...........