Lets calculate molality first
Molar mass of C2H6O2,
MM = 2*MM(C) + 6*MM(H) + 2*MM(O)
= 2*12.01 + 6*1.008 + 2*16.0
= 62.068 g/mol
mass(C2H6O2)= 350.0 g
use:
number of mol of C2H6O2,
n = mass of C2H6O2/molar mass of C2H6O2
=(3.5*10^2 g)/(62.07 g/mol)
= 5.639 mol
m(solvent)= 900.0 g
= 0.9 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(5.639 mol)/(0.9 Kg)
= 6.266 molal
lets now calculate ΔTf
ΔTf = Kf*m
= 1.86*6.2655
= 11.6539 oC
This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 11.6539
= -11.6539 oC
Answer: -11.65 oC
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