Question

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?

Answer 1

`"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...........