Answer
$0.0199\text{ L}$
Work Step by Step
$1.28\text{ g $CaCl_2$}*\frac{1\text{ mol $CaCl_2$}}{110.98\text{ g $CaCl_2$}}*\frac{1\text{ L}}{0.580\text{ mol $CaCl_2$}}=0.0199\text{ L}$
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