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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS