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What is the mass of CaCl2•2H2O (MW 147.02) in grams needed to prepare 1.0E+02 mL of...

What is the mass of CaCl2•2H2O (MW 147.02) in grams needed to prepare 1.0E+02 mL of a 0.4300 M solution of Ca2+ (MW 40.078)?

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Answer #1

MMoles = molarity * volume ( in L )

= 0.4300*1.0*10^2

= 43

CaCl2.2H2O = Ca2+ + 2 Cl- + 2 H2O

MMoles of Ca2+ = moles of CaCl2.2H2O

= 43

Mass = mmoles * molar mass *10^-3

= 43 * 147.02*10^-3

= 6.322 grams

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