Calculate the freezing point: the freezing point of the solution: 50.0 g of glucose, C6H12O6, added to 118 g of water (Kf=1.86°C)

Question

Calculate the freezing point and boiling point of each of the following solutions:

 

1)the freezing point of the solution: 50.0 g of glucose, C6H12O6 , added to 118 g of water (Kf =1.86°C )

 

2)the boiling point of the solution: 50.0 g of glucose, C6H12O6 , added to 118 g of water (Kb =0.52°C )

 

3)the freezing point of the solution: 153 g of sucrose, C12H22O11 , a nonelectrolyte, dissolved in 1.35 kg of water (Kf =1.86°C )

 

4)the boiling point of the solution: 153 g of sucrose, C12H22O11 , a nonelectrolyte, dissolved in 1.35 kg of water (Kb =0.52°C )

Answer 1

1) Δt = i Kf m  , where Δt = depression in freezing point , when solute is being added to water ( t1= 0 degree Celsius ,t2 is freezing point of solution), i= number of ions from after dissociation , Kf= constant, m= molality ( mole of solute/kg of solvent).

now, t1= 0 degree cel, t2 unknown, i=1 (glucose is not ionic, so no dissociation),kf= 1.86 degree cel.,m= mole of glucose/kg of water

mole of glucose = wt of glucose/mol wt of glucose= 50/180= 0.28.

118 g= 118/1000 or 118 *10^-3 kg( since 1 kg= 1000g)

molality = 0.28/(118*10^-3)= 2.37 

put the values,

 

0- t2= 1*1.86*2.37 = 4.4

t2= -4.4 degree cel.

 

 

 

2) ΔT = (i Kb m )  here T2= BP of water,(100 degree cel) ,t1=  bp of solution, i= no of ions after dissociation,  kb= constant,m= Molarity = number of moles of solute/kg of solvent

 

here, t1= unknown, t2= 100, i=1(non electrolyte),kb=0.52, m= 2.37 (as Q=1, everything is same).

put the values,

t1-100=1*0.52*2.37

t1= 101.2 degree cel.

 

3 and 4 are similar type.
all the best!!!