Question

How many grams of solute are present in
845 mL
of
0.730 M KBr?

Answer 1

V(s) = 845 ml = 0.845 l

c(KBr) = 0.730 M

m(KBr) - ?

n(KBr) = c*V =0.617 mol

m(KBr) = 73.4 g