22.39 grams.
Explanation
How many moles of CaCl₂ in that solution?
[tex]c = 0.40 \; \text{M} = 0.40 \; \text{mol}\cdot \textbf{L}^{-1}[/tex];
[tex]V = 504.3 \; \textbf{ml} = 0.5043 \; \textbf{L}[/tex];
[tex]n = c \cdot V = 0.40 \; \text{mol}\cdot \textbf{L}^{-1} \times 0.5043 \; \textbf{L}= 0.20172 \; \text{mol}[/tex].
What's the mass of that 0.20172 moles of CaCl₂?
Molar mass from a modern periodic table:
Molar mass of CaCl₂:
[tex]M(\text{CaCl}_2) = 40.078 + 2 \times 35.45 = 110.978 \; \text{g}\cdot \text{mol}^{-1}[/tex].
Mass of that 0.20172 moles of CaCl₂:
[tex]m = n \cdot M = 0.20172 \times 110.978 = 22.39 \; \text{g}[/tex].
