Question

What is the molality of a solution of 49.8 g of ethanol (CH3CH2OH) in 441 mL of water if the density of water is 1.0 g/mL?

A) 1.80 m B) 0.0353 m C) 1.96 m D) 0.0341 m E) 2.45 m

The answer is 2.45 m but how do you do this?

Answer 1

molality : number of moles of solute present in 1 Kg of solvent

first find moles

moles of solute = weight / molar mass

weight = 49.8 g

molar mass = 46

moles = 49.8 /46

         = 1.083

weight of solvent (water) = density x volume

                                     =   1 x 441

                                      = 441 g

                                      = 0.441 kg

molality = moles of solute / weight of solvent in kg

             = 1.083 / 0.441

             = 2.45 m