Question

How would you prepare 3.5 L of a 0.9M solution of KCl?

Answer 1

$V=3,5L\\ Cm=0,9M\\ M_{KCl}=74\frac{g}{mol}\\\\ C_{m}=\frac{n}{V}\\\\ n=\frac{m}{M}\\\\ C_{m}=\frac{m}{MV} \ \ \ \Rightarrow \ \ \ m=C_{m}MV\\\\ m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g$

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.