What volume of 0.125 M CaCI_2 contains 2.22 times 10^21 Cl^- ions?

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Answer 1

Answer #1

Cl- moles = number of Cl- ions / Avagadro number

= ( 2.22 x 10^21) / ( 6.023 x 10^23 ) = 0.003686

CaCl2 moles = ( 1/2) CaCl2 moles = ( 1/2) (0.003686) = 0.001843 ( since 1CaCl2 gives 2Cl-)

Molarity of CaCl2 = Moles of CaCl2 / Volume of CaCl2

0.125 M = 0.001843 / vol

vol = 0.01474 L = 14.74 ml

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Answer 2

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