Molality question on chemistry

Calculate the volume ( L ) of the solute acetonitrile and the mass ( kg ) of the solvent C6H6 that should be combined to produce 8.81 kg of a solution that is 0.396 m acetonitrile.

work involved would be appreciated. thanks!

molality = moles solute / kg solvent

So a 0.396 m solution has 0.396 moles of ACN in 1.00 kg C6H6

We can use this to determine the % mass Acetonitrile and C6H6 in a solution prepared with 1.00 kg of C6H6, then apply this % to the mass of solution you want to make.

work out the mass of ACN that is mixed with 1.00 kg C6H6
mass = molar mass x moles
= 41.05 g/mol x 0.396 mol
= 16.2558 g

So, we have 16.26 g of ACN mixed in 1.00 kg of C6H6

% mass ACN = mass ACN / total mass x 100/1
= 16.26 g / 1016.26 g x 100/1
= 1.60 %

The solution as prepared will be 1.60 % by mass ACN and 98.40 % C6H6 by mass

So mass C6H6 needed for 8.81 kg of solution
= 0.9840 x 8.81 kg
= 8.67 kg (3 sig figs)

mass ACN = 0.0160 x 8.81 kg
= 0.141 kg
= 141 g

Now we need to convert mass ACN to volume
density ACN = 0.777 g/ml
therefore volume ACN = mass / density
= 141 g / 0.777 g/ml
= 181 ml (3 sig figs)

That's the total mass of the solution. 16.26 g of acetonitrile mixed with the 1 kg (1000 g) of C6H6. Then you go on to use this to determine % mass of acetonitrile and C6H6.

Report Abuse