how many gramof ethylene glycol CH20HCH2OH Must be added to 37.8 g of water to give a freezing point of - 0.150°c
Answers
Answer:The extent of freezing point depression depends only on the solute concentration (molality
concentration) that can be estimated by a simple linear relationship with the cryoscopic
constant: ΔTF = KF · Cm,
Where, ΔTF - the freezing point depression, is defined as TF (pure solvent)-TF (solution)= 0-(-
0.150)=0.150;
KF is the cryoscopic constant, which is dependent on the properties of the solvent, not the
solute. For water, KF = 1.853 K·kg/mol;
Cm - molality concentration (mole per kg of solvent or math equation is:
M m(H O)
m
C
2
m
).
From this equation the molality of solution is: 0.081mol/kg
1.853
0.150
K
T
C
F
F
m
.
From equation of Cm we calculate the mass of solute: m=Cm·M·m(H2O).
The molar mass of CH2OHCH2OH is: M(C2H6O2)=12·2+1·6+16·2=62 g/mol.
The mass of CH2OHCH2OH is: m= 0.081·62·
1000
37.8
=0.19 g.
Answer: The mass of ethylene glycol is 0.19 g.