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12-07-2017
Chemistry

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How many grams of solute are present in 625 mL of 0.520 M KBr?

Respuesta :

dec77cat dec77cat

12-07-2017

M(KBr)=119.0 g/mol
v=625 mL=0.625 L
c=0.520 mol/L

n(KBr)=cv
m(KBr)=n(KBr)M(KBr)

m(KBr)=cvM(KBr)

m(KBr)=0.520*0.625*119.0=38.675 g

38.675 g

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