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Answer to Question #169936 in General Chemistry for reece

Question #169936

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


1
Expert's answer
2021-03-08T07:07:51-0500

M=n/V

n=M*V=0.730*0.845=0.617 mol

Mr(KBr)=39+80=119


m(KBr)=n*M=0.617*119=73,4g


Answer: 73,4g


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