A 25.0g sample of glucose, C6H12O6, dissolves in 525mL of distilled water.
a. what is the boiling point elevation of the solution over the pure solvent?
b. At what temperature will the solution boil at sea level?
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I'm not a fan of these types of questions. This is the work I did for A:
boiling point elevation is
deltaT = i*Kb*m
i=7 kb=.512
moles of solute is 25g/180gpermole = .139 moles of glucose
i believe it's assumed that mL is grams, so 525 mL = 525g
and because (i believe) you need molality of solvent, this is what i have:
mols solvent/kg solvent
= 29mol water/.525kg water = 55.56
and finally, in the actual formula:
deltaT = 7(.512)(55.56)
= 199.11
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For some reason this answer doesn't seem right. What am I doing wrong here? Also, how do you do part B?
Thank you so much!
you have everything right except you molality is moles of solute over kg of solvent so it would be .139/.525 for molality.
I don't know how you got i=7 lol
Way off - do you even know what i is?
I is 1. This is a covalent molecule
To determine the boiling point elevation of the solution, you need to correctly calculate the molality of the solvent and the number of particles (i) in the solute.
Let's start by calculating the molality of the solvent. You correctly identified that 525 mL of water is equivalent to 525 g. However, the molar mass of water is 18 g/mol, not 29 g/mol as you mentioned. So, the number of moles of water is:
moles of water = mass of water / molar mass of water
= 525 g / 18 g/mol
= 29.17 mol
Next, calculate the molality of the solvent (water) using the formula:
molality = moles of solute / mass of solvent (in kg)
Here, the mass of water is 525 g (which equals 0.525 kg). The moles of solute is 0.139 mol.
molality = 0.139 mol / 0.525 kg
= 0.265 mol/kg
Now, let's calculate the boiling point elevation using the formula:
ΔT = i * Kb * m
The van't Hoff factor (i) for glucose (C6H12O6) is 1, as it does not dissociate into separate ions. The boiling point elevation constant (Kb) for water is 0.512 °C/m.
ΔT = 1 * 0.512 °C/m * 0.265 mol/kg
= 0.139 °C
So, the boiling point elevation of the solution over the pure solvent is 0.139 °C.
Now, let's move on to part B, which asks for the temperature at which the solution will boil at sea level.
At sea level, the atmospheric pressure is approximately 1 atm. The boiling point of water at 1 atm is 100 °C.
To calculate the actual boiling point of the solution, you need to consider the boiling point elevation.
T_solution = T_boiling point of pure solvent + ΔT
T_solution = 100 °C + 0.139 °C
= 100.139 °C
Therefore, the solution will boil at approximately 100.139 °C at sea level.