The formula for depression in freezing point is given by:
[tex]\Delta T_{f}=m\times K_{f}[/tex] (1)
where,
[tex]\Delta T_{f}[/tex] = depression in freezing point
[tex]K_{f}[/tex] = freezing point depression constant =[tex]1.86^{o}C kg/mol[/tex]
m = molality
First calculate the molality which is equal to
molality =[tex]\frac{weight of solute in g}{molecular weight of solute \times weight of solvent in kg}[/tex]
= [tex]\frac{w g}{62 \times 1.05 kg}[/tex]
= [tex]\frac{w g}{65.1 kg }[/tex]
Now, [tex]\Delta T_{f}[/tex] = freezing point of water - freezing point of solution
= [tex]0-(-4.66^{o}C)[/tex]
= [tex]4.66^{o}C[/tex]
Now, from formula (1)
[tex]4.66^{o}C= \frac{w g}{65.1 kg}\times 1.86^{o}C kg/mol [/tex]
w in g = [tex]\frac{4.66\times 65.1 }{1.86}[/tex]
= [tex]163.1 g[/tex]
Thus, mass of ethylene glycol = [tex]163.1 g[/tex]
Answer:
163g of ethylene glycol
Explanation:
The addition of a non-volatile solute to a solvent as water produce the decreasing of freezing point following the formula:
ΔT = Kf×m×i
Where ΔT is change in freezing point (0°C- (-4.66°C) = 4.66°C); Kf is freezing point depression constant (1.86°C/m); m is molality (moles of solute / kg solvent); i is Van't Hoff factor (1 for ethylene glycol in water)
Replacing:
4.66°C = 1.86°C/m × moles solute / 1.05kg
2.63 = moles solute.
Molar mass of ethylene glycol is:
C: 12.01×2 = 24.02g/mol
H: 6×1.01 = 6.06g/mol
O: 2×16 = 32g/mol
24.02g/mol + 6.06g/mol + 32g/mol = 62.08g/mol
Thus, mass of ethylene glycol must be added is:
2.63mol × (62.08g/mol) = 163g of ethylene glycol
