How many mL of a 0.100 M solution are required have #1.32 x 10^2# moles of solute?

1 Answer
Dec 1, 2016

Answer:

Rather a large volume, over #1*m^3#

Explanation:

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

And, typically, #"concentration"# has units of #mol*L^-1#.

Here, clearly, #"Volume of solution"="Moles of solute"/"Concentration"#

#=(1.32xx10^2*mol)/(0.100*mol*L^-1)=1320*L#