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Step 1: List the given values.
[tex]\begin{aligned} & M = 4.00 \: M = \text{4.00 mol/L} \\ & MM_{\text{solute}} = \text{129.83 g/mol} \\ & V_{\text{solution}} = \text{650.00 mL = 0.65000 L} \end{aligned}[/tex]
Step 2: Calculate the number of moles of solute.
[tex]\begin{aligned} n_{\text{solute}} & = (M)(V_{\text{solution}}) \\ & = \text{(4.00 mol/L)(0.65000 L)} \\ & = \text{2.60 mol} \end{aligned}[/tex]
Step 3: Calculate the mass of solute.
[tex]\begin{aligned} mass_{\text{solute}} & = (n_{\text{solute}})(MM_{\text{solute}}) \\ & = \text{(2.60 mol)(129.83 g/mol)} \\ & = \text{337.558 g} \\ & \approx \boxed{\text{338 g}} \end{aligned}[/tex]
Hence, 338 g of anhydrous CoCl₂ would be needed.
[tex]\\[/tex]
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