Answer:
1.4776 g
Explanation:
First, let us determine the number of moles present in 100 mL of 0.1005 M solution of Ca2+.
mole = molarity x volume in Liter
mole of Ca2+ = 0.1005 x 0.1 = 0.01005
Also,
mole = mass/molar mass
mass of 0.01005 mole Ca2+ = mole x molar mass
= 0.01005 x 147.02 = 1.4776 g
Since 1 mole of Ca2+ is present is 1 mole of CaCl2·2H2O, 1.4776 g of CaCl2·2H2O would be weighed out in order to prepare 00mL of a 0.1005M solution of Ca2+.