How many gram of solute should be dissolved in 500 g of water so as to get a solution having difference of 105 C between freezing point and boiling point ? Molar mass of solute=342 (Kf=1.86 K...
How many gram of solute should be dissolved in 500 g of water so as to get a solution having difference of 105 C between freezing point and boiling point ?
Molar mass of solute=342
(Kf=1.86 K Kg /mol,Kb=0.52 K Kg /mol for water)
please give the correct step in a order and all the corresponding values of each formula.
The boiling point needs to be raised an amount of delta Tb while the freezing point needs to be lowered an amoung of delta Tf. Since pure water already has a difference of 100 deg C the amount you lower the freezing point plus the amount you raise the boiling point have to add up to total change of 5 degrees C which is also a change of 5 K.
5 K = delta Tf + delta Tb
delta Tf = Kf * m delta Tb = Kb * m
where m is the molality of the solution
molality = n / mass = moles of solute (n) / kg of solvent (mass)
substituting these all into one equation:
5 K = Kf * n/mass + Kb * n/mass
since they both use the same moles of solute and mass of solvent
5 K = (Kf + Kb) * n/mass
solving for n
n = ( 5 K * mass ) / ( Kf + Kb )
n = ( 5 K * 0.5 kg ) / ( 1.86 K kg/mol + 0.52 K kg/mol )
n = 1.05 moles solute
1.05 moles * 342 g/mole = 359 grams of solute (yikes)