What is the molality of a solution produced using 15.2 g of Calcium chloride and 345 g of methanol (CH3OH)?

Without doing any research - I guess that this solution is possible :

Mass of CuCl2 in 1000 g methanol = 1000 g / 345 g * 15.2 g CuCl2 = 44.05 g CuCl2

Molar mass CuCl2 = 63.55 + 2*35.45 = 134.45 g

Mol CuCl2 in 44.05 g 44.05 g / 134.45 g/mol = 0.328 mol dissolved in 1.0 kg CH3OH

Molality = 0.328 m.