Solution:
"\\nu(CaCl_2)=V\\times C_M=0.5 L\\times 6.5 \\frac{mol}{L}=3.25\\,mol"
"m(CaCl_2)=\\nu\\times M=3.25\\,mol\\times111.08\\, \\frac{g}{mol}=361.01 \\,g"
Answer: 361.01 g
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