What molar quantity of potassium bromide is present in a #68.5*mL# volume of a solution whose concentration is #1.65*mol*L^-1# with respect to the salt?

1 Answer
Aug 11, 2016

Answer:

Approx. #0.1# #mol# #KBr#

Explanation:

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

Thus #"moles of KBr"# #=# #68.5xx10^-3cancelLxx1.65*mol*cancel(L^-1)# #=# #0.113*mol#.

I assume (perhaps wrongly!) that you quoted a #"molar"# concentration (i.e. #mol*L^-1#) with respect to #KBr#. You might have specified a #"molal concentration"#, i.e. #"moles of solute"/"kilogram of solvent"#, which is what "#"m"#" generally signifies. The answers would be similar.