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Answer to Question #162774 in General Chemistry for Bella winter
Question #162774
What volume of 0.300 M calcium chloride contains 3.50g of calcium chloride
Expert's answer
3.50g CaCl2 × (1mol CaCl2 /110.98g CaCl2)
= 0.0315 moles
Molarity = no. of moles / volume
0.300 = 0.0315 / volume
Volume = 0.0315/0.300
Volume = 0.105 L
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