Question

Calculate the molality of a solution formed by adding 3.331 g of methanol (CH3OH) to 100.0 mL of water. The density of water is 0.9982 g/mL.

Answer 1

Answer:

Given the mass of methanol=3.331 g,

and the molar mass of methanol=32.04 g/mol.

Therefore moles of methanol=mass/molar mass=3.331 g/32.04 g/mol=0.10396 mol.

Given the volume of water=100 mL and the density =0.9982 g/mL

Mass of water=density x volume=0.9982 g/mL x 100 mL=99.82 g=0.09982 Kg.

Molality=moles of the solute/mass of solvent(Kg)

Here solute is methanol and solvent is water.

Molality=0.10396 mol/0.09982 Kg=1.0415 mol/Kg

Molality=1.0415 m.

Thanks and I hope you like it.