A solution containing equal masses of glycerol, C3H8O3, and water has a density of 1.10 g/mL. Calculate the molality of glycerol.
I got 11.9 m.
Step 1:
Make the assumption that glycerol and water has 50g since they have equal masses
Step 2:
Convert grams of glycerol to moles. If you don't know how to do that then I'd quit chem right now...
moles of C3H8O3=.543
Step 3:
convert the 50g of water into kg.
kg water= .05
Step 4:
Set up molality equation:
(.543)/(.05)= 10.9m
Answer: 10.9m
Well, congratulations on doing the math! But let me ask you this, do you know why the mole of Halloween candy went to the party alone? Because it didn't want to share its molarity with anyone!
Now, let's get back to your question. To calculate the molality of glycerol, we need to determine the number of moles of glycerol dissolved in 1 kilogram of water.
First, we need to find the molar mass of glycerol. By using some chemical trickery (looking at the periodic table), we find that the molar mass of C3H8O3 is approximately 92.09 g/mol.
Next, let's look at our solution. We know that its density is 1.10 g/mL. Since we have equal masses of glycerol and water, we can say that the density of the solution is equivalent to the average density of glycerol and water combined.
So, we can set up an equation:
(1.10 g/mL) = (mass of glycerol) / (volume of glycerol) + (mass of water) / (volume of water)
Since we have equal masses, the equation simplifies to:
1.10 g/mL = (2 * mass of glycerol) / (2 * volume of glycerol)
Simplifying further:
1.10 g/mL = mass of glycerol / volume of glycerol
Now, since density is mass/volume, we can replace mass/volume with density:
1.10 g/mL = density of glycerol
Therefore, the molality of glycerol is 1.10 m.
Hope that clears things up while also adding a little humor to the mix!
To calculate the molality of glycerol, we need to determine the number of moles of glycerol present in the solution.
First, we need to determine the masses of glycerol and water in the solution. Since the solution contains equal masses of glycerol and water, we can assume that each component constitutes 50% of the total mass.
Let's assume we have a total mass of 100 g, so glycerol and water would each have a mass of 50 g.
Next, we need to convert the masses of glycerol and water into moles. To do this, we divide the mass by the molar mass of each component.
The molar mass of glycerol (C3H8O3) can be calculated by adding up the atomic masses of its constituent elements:
(3 x atomic mass of carbon) + (8 x atomic mass of hydrogen) + (3 x atomic mass of oxygen)
= (3 x 12.01 g/mol) + (8 x 1.01 g/mol) + (3 x 16.00 g/mol)
= 92.09 g/mol
To calculate the number of moles of glycerol, we divide the mass of glycerol by its molar mass:
Number of moles of glycerol = mass of glycerol / molar mass of glycerol
= 50 g / 92.09 g/mol
= 0.543 mol
The molality (m) of glycerol is defined as the number of moles of glycerol divided by the mass of the solvent (water) in kg.
To calculate the molality, we convert the mass of water from grams to kilograms:
Mass of water = 50 g = 0.01 kg (since 1 kg = 1000 g)
Molality of glycerol = number of moles of glycerol / mass of water in kg
= 0.543 mol / 0.01 kg
= 54.3 mol/kg
Therefore, the molality of glycerol in the given solution is 54.3 mol/kg.
To calculate the molality (m) of glycerol in a solution, we need to know the molar mass of glycerol (C3H8O3) and the density of the solution.
The molar mass of glycerol can be calculated by adding up the atomic masses of its constituent elements. Carbon (C) has an atomic mass of 12.01 g/mol, hydrogen (H) has an atomic mass of 1.01 g/mol, and oxygen (O) has an atomic mass of 16.00 g/mol. Glycerol has 3 carbon atoms, 8 hydrogen atoms, and 3 oxygen atoms.
Molar mass of glycerol (C3H8O3):
(3 x atomic mass of C) + (8 x atomic mass of H) + (3 x atomic mass of O)
= (3 x 12.01 g/mol) + (8 x 1.01 g/mol) + (3 x 16.00 g/mol)
= 36.03 g/mol + 8.08 g/mol + 48.00 g/mol
= 92.11 g/mol
Now we can calculate the molality of glycerol using the formula:
molality (m) = moles of solute / mass of solvent (in kg)
Since the solution contains equal masses of glycerol and water, we can assume that the mass of solvent is half that of the solution mass. Therefore, the mass of solvent is equal to the mass of the solution divided by 2.
Given that the density of the solution is 1.10 g/mL, we can calculate the mass of the solution as follows:
mass of solution = volume of solution x density
mass of solution = 1000 mL x 1.10 g/mL (since density is given in g/mL)
mass of solution = 1100 g
mass of solvent = mass of solution / 2
mass of solvent = 1100 g / 2
mass of solvent = 550 g
To find the moles of glycerol, we divide the mass of glycerol by its molar mass:
moles of glycerol = mass of glycerol / molar mass of glycerol
moles of glycerol = 550 g / 92.11 g/mol
moles of glycerol ≈ 5.976 mol (rounded to 3 decimal places)
Finally, we can calculate the molality of glycerol:
molality (m) = moles of glycerol / mass of solvent (in kg)
molality (m) = 5.976 mol / 0.550 kg
molality (m) ≈ 10.87 m (rounded to 2 decimal places)
Therefore, the molality of glycerol in the solution is approximately 10.87 m.
Assume 1000ml of the solution.
total mass=1000ml/density=1100g
moles of glycerol=550/molmassgly /1L=550/92.1= not your answer.