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...........

Related questions
See all questions in Molarity
Impact of this question
17937 views around the world
You can reuse this answer
Creative Commons License