Question

calculate the difference in boiling point for each of the three mixtures

1. MeOAc+EtOAc

2. MeOAc+PrOAc

3. MeOAc+ BuOAc

Answer 1

The difference in boiling point of solvent mixture calculated as :

Difference in boiling point ( boiling point elevation temp.) = ΔKb*M * ΔTb

where Kb = Elullioscopic constant , M =molality., Tb= Boiling point

Hence

ΔKb ( MeOAc + EtOAc) = Kb(EtOAc) - Kb(MeOAc)

= 2.82 - 2.21

= 0.61

So,  

1. Difference in boiling point of MeOAc + EtoAc = [ Tb of EtOAc - Tb of MeOAc ] * ΔKb*M

= (77.1 - 57.1)* 0.61*1

= 12.2 degree celsius

Similarly,

2. Difference in boiling point of MeOAc + ProAc = (102 - 57.1 )*0.79*1

= 35.47 degee celsius

3. Difference in boiling point of MeOAc + BuOAc = (126 -57.1 )* 0.91*1

= 62.69 degree celsius