1. How many grams of CaCl2 would be required to produce a 8.5 M (molar) solution with a volume of 2.0 L?
The molar mass of CaCl2 is 40.078 + 2*35.45 = 111 g/mol.
For 2.0 L of 8.5 M solution it requires 2.0*8.5*111 = 1887 g of CaCl2 or 1900 g (round off to two significant figures).
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