1. how many grams of cacl2 would be required to produce a 3.5 m (molar) solution with a volume of 2.0 l? . 2. what is the molarity of a 5.00 x 102 ml solution containing 2490 g of ki? . 3. how many moles of lif would be required to produce a 2.5 m solution with a volume of 10.5 l? .

Explanation:

1.) Moles of calcium chloride = n

Molarity of the solution = 3.5 M

Volume of the solution =V= 2.0 L

n = 7 moles

Mass of 7 moles of calcium chloride :

777 grams of calcium chloride would be required.

2.) Mass of the KI = 2490 g

Moles of KI = n =

Volume of the solution,V =

Molarity of the KI solution is 30 mol/L.

3.)Molarity of the LiF solution = 2.5 M

Volume of the solution = 10.5 L

Moles of LiF = n

n = 26.25 moles

26.25 moles of LiF would be required.

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