How do you convert 11.0 g #CH_3OH# to 0.490 M?

1 Answer
Dec 1, 2016

Answer:

You want to make a methanolic solution in water whose concentration with respect to methanol is #0.490*mol*L^-1#?

Explanation:

And thus #"Moles of methanol"/"Volume of solution"# #=# #0.490*mol*L^-1#

#"Methanol"# has a formula mass of #32.04*g*mol^-1#.

And thus #((11.0*g)/(32.04*g*mol^-1))/"Volume of solution"=0.490*mol*L^-1#.

Upon rearrangement, #"Volume of solution"=((11.0*g)/(32.04*g*mol^-1))/(0.490*mol*L^-1)#, which is under #1*L#.

Is this answer dimensionally consistent? Does it give an answer in #L#, which is required?