Calculate the mass, in grams, of cucl2 (mw = 134.452 g/mol) required to prepare 250.0 ml of a 6.11 % w/v cu2 (mw = 63.546 g/mol) solution.

6.11% w/v of Cu2 + implies that 6.11 g of Cu2 + is present in 100 ml of the solution

therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g

# moles of Cu2 + = 15.275 g/63.546 g mole-1 = 0.2404 moles

1 mole of CuCl2 contain 1 mole of Cu2 + ion

Hence, 0.2404 moles of Cu2 + would correspond to 0.2404 moles of CuCl2

Molar mass of CuCl2 = 134.452 g/mole

The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams