Question

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

A. Add 23 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.

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.

C. Add 567 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.

D. Add 287 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.

Answer 1

`V=3,5L\\ Cm=0,9M\\ M_(KCl)=74(g)/(mol)\\\\ C_(m)=(n)/(V)\\\\ n=(m)/(M)\\\\ C_(m)=(m)/(MV) \ \ \ \Rightarrow \ \ \ m=C_(m)MV\\\\ m=0,9\grac{mol}{L}*74(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.