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Answer:
17.54 g
Explanation:
the freezing point of the solution to -8.8°C
depression in freezing point = 8.8°C
Molal freezing point depression constant of ethylene glycol Kf = 3.11⁰C /m
ΔTf = Kf x m , m is no of moles of solute per kg of solvent .
Let gram of ethylene glycol required be x .
moles = m / mol weight of ethylene glycol
= x / 62
100 g of water = .1 kg
moles of solute dissolved in 1 kg of water
m = x / 62 x .1
m = 10 x / 62
Using the above equation for depression in freezing point
8.8 = 3.11 x 10 x / 62
x = 17.54 g .
The mass of ethylene glycol required to lower the freezing point is 17.54 g.
The given parameters:
Molar freezing point depression constant of ethylene glycol Kf = 3.11⁰C /m
Let the mass of ethylene glycol required be x .
The number of moles of the ethylene is calculated as follows;
[tex]n = \frac{x}{M} \\\\n = \frac{x}{62 \times 100 g \ of H_2 O}\\\\n = \frac{x}{6.2}[/tex]
Apply the following equation for depression in freezing point as follows;
[tex]\Delta T_f = n Kf\\\\8.8 = 3.11 \times \frac{x}{6.2} \\\\3.11 x = 54.56\\\\x = \frac{54.56}{3.11} \\\\x = 17.54 \ g[/tex]
Thus, the mass of ethylene glycol required to lower the freezing point is 17.54 g.
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