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?

Question

Noah Aldrich · Accepted Answer

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

Here, #"concentration"~=0.4*mol*L^-1# 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.# As an alternative, we may employ the antiquated phrase. #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...........

Jose Ayala · Answer

undefined To calculate the molarity of the diluted solution, you can use the formula:

M1V1 = M2V2

Where:
- M1 is the initial molarity of the solution (0.743 M)
- V1 is the initial volume of the solution (87.44 mL)
- M2 is the final molarity of the solution (what we want to find)
- V2 is the final volume of the solution after dilution (150.00 mL)

Rearranging the formula to solve for M2:

M2 = (M1 * V1) / V2

Substituting the given values:

M2 = (0.743 M * 87.44 mL) / 150.00 mL

M2 ≈ 0.431 M

Therefore, the molarity of the diluted solution is approximately 0.431 M.