Chemistry Problem?

Question

Calculate the boiling point and freezing point of the follwoing solutions of molecular substances:

97.5 g of C12H22O11 (sucrose) in 185 g of water

Answer 1

the molecular weight of sucrose is 342 g/mol.
To find the number of moles:
97.5/342 = 0.285 moles
to find molality
0.285/0.185 = 1.54 molal

since sucrose does not dissosiate, we will not need to calculate the amount of ions in the solution - the molality of particles is the same.

For boiling:

ΔTb = (Kb)(m)
where, accoding the the litature, Kb for water is 0.512 °C/m
=(0.512 °C/m )(1.54 m)=0.788 °C
100 °C + 0.788 °C = 100.788 °C

For freezing:

ΔTf = (Kf)(m)
Where the Kf for water is 1.86 °C/m
=(1.86 °C/m)(1.54 m)
=2.86 °C
0°C - 2.86 °C = -2.86 °C


There you go! ☺

Answer 2

moleclar weight of saccharose
MW =12*12+22+11*16=342g
concentration of the solution 97.5/0.185 =527g/l
molarity of the solution = 527/342=1.54mole/L
As saccharose is non-electrolyte than 1.54mole/L =1.54Osm/L

Ebullioscopy elevation of boiling point 1.54*0.512=0.79°C
Boiling point 100.79°C

freeezing point depression= 1.86*1.54 =2.87°C
freezing point = -2.87°c