Answer
62.5 mL of that solution contains 0.0500 mole of $Ca(NO_3)_2$
Work Step by Step
1. Since the molarity of that solution is equal to $ 0.800 \space M \space Ca(NO_3)_2$:
$1 \space L \space solution = 0.800 \space moles \space Ca(NO_3)_2$
$\frac{1 \space L \space solution}{ 0.800 \space moles \space Ca(NO_3)_2} $ and $\frac{ 0.800 \space moles \space Ca(NO_3)_2}{1 \space L \space solution}$
2. To convert from milliliters to liters: 1 L = 1000 mL
$\frac{1 \space L}{1000 \space mL}$ and $\frac{1000 \space mL}{1 \space L}$
3. Use the conversion factors to calculate the volume (in milliliters) of that solution that has 0.0500 mole $KBr$:
$0.0500 \space mole \space Ca(NO_3)_2 \times \frac{1 \space L \space solution}{ 0.800 \space moles \space Ca(NO_3)_2} \times \frac{1000 \space mL}{1 \space L} = 62.5 \space mL \space solution$
