Answer to Question #169936 in General Chemistry for reece

General Chemistry

Question #169936

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

Expert's answer

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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