Answer to Question #92001 in General Chemistry for jimbo

Question #92001
How many grams of solute are present in
845 mL
of
0.730 M KBr?
1
Expert's answer
2019-07-26T03:16:56-0400

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


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