Answer to Question #99862 in General Chemistry for Cookie

Question

How many grams of solute are present in 845ml of 0.670 M KBr?

Answer 1

@$c=\frac{n}{V}@$

@$n=c\times V @$

@$n(KBr) = 0.670\times0.845 = 0.5662 mol@$

@$m(KBr)= M\times n = 119.00 \times 0.5662 = 67.37 g@$