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?

1 Answer
anor277
Jun 30, 2017

"Concentration"="Moles of solute"/"Volume of solution"

Here, "concentration"~=0.4*mol*L^-1

Explanation:

And we have initially, 87.44xx10^-3*Lxx0.743*mol*L^-1=0.0650*mol, 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_1V_1=C_2V_2, i.e.

C_2=(C_1V_1)/(V_2)=(0.743*mol*L^(-1)xx87.44*cancel(mL))/(150.00*cancel(mL))

Which of course is the same quotient.........which explicity gives an answer with units of mol*L^-1, as required...........

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