How many grams of solute are present in 935 mL of 0.720 M #KBr#?

1 Answer
Jul 10, 2016

Answer:

#"Grams of KCl"# #~=# #50*g#

Explanation:

#"Concentration"="Moles of solute"/"Volume of solution"#. This gives units of #mol*L^-1#.

Thus #"moles of solute"="volume"xx"concentration".=#

#935*cancel(mL)xx10^-3cancelL*cancel(mL^-1)xx0.729*cancel(mol)*cancel(L^-1)xx74.55*g*cancel(mol^-1)#

#=# #??g#