A solution was prepared by dissolving 18.00 grams glucose in 150.0 grams water. The resulting solution was found to have a boiling point of 100.34 0C. Calculate the molar mass of glucose. (Kb (water) = 0.51 0C kg/mol; Tb(water) = 80. 0 0C)
I'm not sure what you mean for Tb = 80. If you mean boiling point for H2O that is 100 C and that's the number I will use.
delta T = Kb*molality
100.34-100 = 0.34 =*0.51 molality.
Solve for molality = mols/Kg solvent.
You know m and you know kg solvent is 0.150. Solve for mols.
Then mols = grams/molar mass. You know mols and grams, solve for molar mass = ?
Post your work if you get stuck. I ran through the calculations in my head and got about 180 g/mol.
Nothing
yes
Well, well, well! Looks like we have a boiling point problem on our hands. But don't you worry, I'm here to help you crunch those numbers and add a little humor to the mix!
To find the molar mass of glucose, we can use the formula:
ΔTb = Kb × m × i
Here, ΔTb is the change in boiling point, Kb is the molal boiling point constant for water, m is the molality of the solution, and i is the van't Hoff factor.
First, let's calculate the molality (m):
m = moles of solute / mass of solvent in kg
The mass of water in kg can be calculated by dividing the mass of water in grams by 1000:
mass of water (kg) = 150.0 g / 1000 = 0.150 kg
If we assume the solution is ideal and nonvolatile, then the molality is equal to the mole fraction of the solute:
m = 18.00 g / (18.00 g + 0.150 kg) ≈ 0.1056 m
Now, let's calculate the change in boiling point (ΔTb):
ΔTb = 100.34 ⁰C - 100.0 ⁰C = 0.34 ⁰C
Now, we can plug in the values into the formula and solve for the molar mass (M):
ΔTb = Kb × m × i
0.34 ⁰C = 0.51 ⁰C kg/mol × 0.1056 m × i
Simplifying the equation, we get:
i ≈ 0.34 ⁰C / (0.51 ⁰C kg/mol × 0.1056 m)
i ≈ 6.53
Since glucose is a non-electrolyte, its van't Hoff factor (i) is approximately 1. So, we can rewrite the equation as:
1 ≈ 0.34 ⁰C / (0.51 ⁰C kg/mol × 0.1056 m)
1 ≈ 0.34 ⁰C / (0.0547056 ⁰C kg/mol)
1 ≈ 6.2156
Now, to find the molar mass (M):
M = mass / moles
The mass of glucose is 18.00 grams, and we can calculate the moles using the molality (m):
moles = m × mass of solvent in kg
moles = 0.1056 m × 0.150 kg
moles ≈ 0.01584 mol
Finally, let's calculate the molar mass:
M ≈ 18.00 g / 0.01584 mol
M ≈ 1137.99 g/mol
So, the molar mass of glucose is approximately 1137.99 grams per mole.
And there you have it! The numbers have been crunched, and the humor has been added. I hope this solution brings a smile to your face! Just remember, glucose may be sweet, but chemistry can make you feel a little sour sometimes. Keep smiling!
To calculate the molar mass of glucose, we need to use the formula:
∆Tb = Kb * m * i
where ∆Tb is the boiling point elevation, Kb is the molal boiling point elevation constant for water, m is the molality of the solute, and i is the van't Hoff factor.
First, let's calculate molality (m) using the given data:
m = moles of solute / mass of solvent in kg
The mass of solvent (water) is given as 150.0 grams. To convert it to kg, divide by 1000:
mass of solvent = 150.0 g / 1000 = 0.150 kg
The moles of solute (glucose) can be calculated using its mass and molar mass. Let's assume the molar mass of glucose as M:
moles of solute = mass of solute / molar mass of solute
Substituting the given values:
18.00 g / M = moles of solute
Now, let's calculate ∆Tb using the boiling point of the solution and the boiling point of pure water:
∆Tb = Tb(solution) - Tb(pure water)
∆Tb = 100.34 0C - 100.00 0C = 0.34 0C
Next, we need to calculate the van't Hoff factor (i), which represents the number of particles formed per molecule of solute. For glucose, it is assumed to be 1.
Now, let's substitute the known values into the formula:
0.34 0C = 0.51 0C kg/mol * m * 1
Solve for molality (m):
m = 0.34 0C / (0.51 0C kg/mol) = 0.6667 mol/kg
Finally, substitute the calculated molality into the moles equation to solve for the molar mass of glucose (M):
18.00 g / M = 0.6667 mol/kg
Rearrange the equation to solve for M:
M = 18.00 g / (0.6667 mol/kg)
M ≈ 27.00 g/mol
Therefore, the molar mass of glucose is approximately 27.00 g/mol.