# 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.

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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)