Question

61 La 60 Nd 140 008 144 242 Praseodymium Neodymium Pr Pm Sm 138.905 Lanthanum 140 116 Cerium 144.913 Prometh 150.36 90 94 AC Th Ра U Np Pu 227020 Actinum 232 038 2310 238.029 2370 244064 How many formula units are contained in 63.0 g of CaC Select the correct answer below: O 3.42 x 1023 formula units O 3.54 x 1023 formula units O 1.75 x 1023 formula units O 3.19 x 1023 formula units

Answer 1

1 mole of any substance has 6.023 × 10^23 particles.

1 mole of CaCl2 = 6.023 × 10^23 formula units.

Number of moles = mass/molar mass

Molar mass of CaCl2 = 110.98 g/mol

Moles of CaCl2 = 63/110.98

Moles of CaCl2 = 0.568 moles

Number of formula units = 0.568 × 6.023 × 10^23

= 3.42 × 10^23 formula units.