what mass of solute is needed to prepare 400 mL of 0.850 M CuSO4 from CuSO4.5H2O
No of moles= molarity×volume
=0.85×0.4
=0.34moles
RFM of CuSO4= 64+32+4×16= 160
mass=no of moles×RFM
=0.34×160
=54.4g
NOTE: That's 54.4 g CuSO4. But the problem asks for grams of CuSO4.5H2O so that is
54.4 g CuSO4 x (molar mass CuSO4.5H2O/molar mass CuSO4) = 54.4 x 250/160 = 85 g.