For the purpose we have to know following equation for freezing
point depression:
ΔTf=Kf*b
and that Kf of water is 1.86°C/m.
From the formula we can determine molality (b):
b=Δtf/Kf=0.77/1.86= 0.414 mole/kg
Now when we know molality, we can calculate mass of sucrose
in solution:
b=n/m(H20) => n(sucrose)=b*m(H20) = 0.414 mole/kg *0.1 kg = 0.0414 mole
Finally, we can determine the molar mass of sucrose:
M=m/n=14.2/0.0414= 343g/mole
