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Answer :
The boiling point of the mixture is approximately 248.48 °C.
To calculate the boiling point of the mixture, we need to use the formula for boiling point elevation. The formula is: ΔTb = Kb * m * i
In this case, the boiling point elevation constant for H2O (Kb) is given as 0.512 "Chm. The mass of the ethylene glycol (m) is 95.0 g, and the mass of water (H2O) is 195 g.
The "i" in the formula represents the van't Hoff factor, which is the number of particles that the solute dissociates into in the solvent. In this case, ethylene glycol does not dissociate in water, so the van't Hoff factor (i) is 1.
Substituting the values into the formula, we get: ΔTb = 0.512 * (95.0 + 195) * 1
Calculating this gives us: ΔTb = 0.512 * 290
ΔTb = 148.48
The boiling point elevation (ΔTb) is 148.48 °C.
To find the boiling point of the mixture, we need to add this to the boiling point of pure water, which is 100 °C.
Boiling point of the mixture = 100 + 148.48 = 248.48 °C
Since none of the answer options match exactly, it seems there might be an error in the given choices.
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The boiling point of the mixture is 104 °C and in order to determine it, we need to consider the boiling point elevation caused by the presence of solute, ethylene glycol [tex](HOCH_{2} CH_{2}OH)[/tex], in water [tex](H_{2} O)[/tex].
The boiling point elevation can be written as:
ΔT = [tex]K_b * m[/tex]
where ΔT is the boiling point elevation, [tex]K_b[/tex] is B.P. elevation constant, and m is molality of solute.
First, let's calculate the molality (m) of the ethylene glycol solution:
Number of moles of ethylene glycol [tex](HOCH_{2}CH_{2} OH)[/tex]:
The molar mass of [tex](HOCH_{2}CH_{2} OH)[/tex] = 62.07 g/mol
Moles of [tex](HOCH_{2}CH_{2} OH)[/tex]= mass / molar mass = 95.0 g / 62.07 g/mol
Calculate the mass of water (H2O) in kilograms:
Mass of water = 195 g
Mass of water in kg = 195 g / 1000 g/kg
Calculate the molality (m):
Molality (m) = moles of [tex](HOCH_{2}CH_{2} OH)[/tex] / mass of water (in kg) = (95.0 g / 62.07 g/mol) / (195 g / 1000 g/kg)
Next, we can calculate the boiling point elevation (ΔT):
Boiling point elevation constant [tex](K_b)[/tex] = 0.512 °C/m
ΔT =[tex](K_b)*m[/tex]
Substituting the values:
ΔT = 0.512 °C/m × [(95.0 g / 62.07 g/mol) / (195 g / 1000 g/kg)]
ΔT = 0.512 °C/m × [(1.53 mol) / (0.195 mol)]
ΔT = 0.512 °C/m × (7.846)
ΔT = 4 °C
To find the boiling point of the mixture, we need to add the boiling point elevation (ΔT) to the boiling point of pure water, which is 100 °C.
Boiling point of mixture = 100 °C + ΔT
= 100 °C + 4°C
=104 °C
Hence, option C, i.e. 104 °C is the correct answer.
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