A hydrate of copper (II) chloride, CuCl2∙nH2O, is heated until the mass of the anhydrate remains constant at 19.72 g, indicating a loss of 5.32 g of water. How many moles of water are present for each mole of salt in the original hydrate? (Find n.)

n = 2

Explanation:

To solve this question we must find the moles of the anhydrate CuCl2 (Obtained from the mass of 19.72g) and the moles of water (Obtained from the 5.32g):

Moles CuCl2 -Molar mass: 134.45g/mol-

19.72g * (1mol / 134.45g) = 0.14667 moles

Moles H2O -Molar mass 18.01g/mol-

5.32g * (1mol / 18.01g) = 0.2954 moles

The ratio of moles of H2O:CuCl2 is:

0.2954 moles / 0.14667moles = 2.0

That means n = 2 and the formula is:

CuCl2∙2H2O

14 days is the half-life of phosphorus-32

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.