Question

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

Answer 1

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

$n=c\times V$

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

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