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