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

1 answer:

4 0

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

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.

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