what volume, in liters, of 1.5 m cacl2 solution can be made using 1200.0 g cacl2? cacl2: 110.98 g/mol

Answer:

7.23 L

Explanation:

Step1: Calculate moles of CaCl₂

$n=\frac{m}{M}=\frac{1200.0\ g}{110.98\ g/mol}\approx10.81\ mol$

Step2: Use molarity formula to find volume

$V = \frac{n}{M}=\frac{10.81\ mol}{1.5\ mol/L}\approx7.23\ L$