Calculate the molality of a solution that is prepared by mixing 25.5 ml of ch3oh (d = 0.792 g/ml and 387 ml of ch3ch2ch2oh (d = 0.811 g/ml.

The solvent is ch3ch2ch2oh with a volume of 387 mL which is equivalent to

387 mL (0.811 g/mL) (1 kg / 1000 kg) = 0.3139 kg

The moles of the alchol is

25.5 mL (0.792 g/mL) (1 mol/32 g) = 0.631 mol

The molality is

0.631 mol / 0.3139 kg = 2.01 mol/kg