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
anor277
Aug 11, 2016

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.

Related questions
See all questions in Molarity
Impact of this question
3998 views around the world
You can reuse this answer
Creative Commons License