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.
Oh, so you're diving into the world of molality, huh? Well, let's calculate this together, but I must warn you, I tend to make chemical solutions funny, not boring.
To calculate the molality (m), we need to use the formula:
m = (moles of solute) / (mass of solvent in kg)
First, let's convert the given density of 1.10 g/mL to kg/L because it's easier for us. So, 1.10 g/mL is equal to 1.10 kg/L.
Now, let's assume we have 1 L of this solution. Since the densities of glycerol (C3H8O3) and water are known, we can find the mass of each component using their respective densities:
Let's assign the mass of glycerol as m1 and the mass of water as m2.
1.10 kg/L = (m1 + m2) kg
Since we are told that the solution contains equal masses of glycerol and water:
m1 kg = m2 kg
So, we can rewrite the equation as:
1.10 kg/L = (m1 + m1) kg
Simplifying it further:
1.10 kg/L = 2m1 kg
Now, let's solve for m1 (mass of glycerol):
m1 = (1.10 kg/L) / 2
m1 = 0.55 kg/L
Great, now we know the mass of glycerol for 1 L of solution. But our question asks for molality, which needs the mass of the solvent in kg.
Since the mass of glycerol is equal to the mass of water, the mass of the solvent is 2 x m1:
Mass of solvent = 2 x 0.55 kg/L
Mass of solvent = 1.10 kg/L
Finally, we can find the molality of glycerol:
m = (moles of solute) / (mass of solvent in kg)
But wait, we're not given the moles of solute (glycerol) or the molar mass of glycerol. So, I'm sorry, but I can't calculate the molality for you with the given information.
But hey, I hope my attempt at humor made this chemical calculation a little more enjoyable for you!
To calculate the molality of glycerol in a solution, we need to know the mass of glycerol and the mass of the solvent (water, in this case).
Given:
Density of the solution = 1.10 g/mL
Mass of the solution = Mass of glycerol + Mass of water
Let's assume that we have 1 g of the solution
Therefore, the volume of the solution would be:
1 g / 1.10 g/mL = 0.909 mL
Since the solution contains equal masses of glycerol and water, each component would have a mass of 0.5 g (half of the total mass).
To calculate the molality, we need to convert the mass of glycerol to moles and divide it by the mass of the solvent in kilograms.
Molar mass of glycerol (C3H8O3) = 92.09 g/mol
Number of moles of glycerol = mass of glycerol / molar mass of glycerol
Number of moles of glycerol = 0.5 g / 92.09 g/mol = 0.005428 mol
Mass of water = mass of solvent = 0.5 g
To convert grams to kilograms, divide by 1000:
Mass of water = 0.5 g / 1000 = 0.0005 kg
Molality = moles of glycerol / mass of water in kg
Molality = 0.005428 mol / 0.0005 kg = 10.856 mol/kg
Rounded to three significant figures, the molality of glycerol in the solution is approximately 10.9 m, not 11.9 m.
To calculate the molality of glycerol in the solution, we first need to determine the mass of glycerol present in the solution.
Given:
Density = 1.10 g/mL
Mass of the solution = mass of glycerol + mass of water
Since equal masses of glycerol and water are present, we can assume that the mass of glycerol is equal to the mass of water.
Let's assume that the mass of glycerol and water is 'm' grams each.
So, the total mass of the solution would be 2m grams.
Given that the density of the solution is 1.10 g/mL, we can write:
Density = Mass / Volume
1.10 g/mL = (2m g) / (2 mL)
Now, let's calculate the mass of 1 mL of the solution:
1.10 g/mL = (2m g) / (2 mL)
1.10 g/mL = m g/mL
Now, we know that the density of pure water at room temperature is approximately 1 g/mL. Since the mass of water in the solution is the same as the mass of glycerol, the mass of 1 mL of the solution is equivalent to 1 g.
Therefore:
m g/mL = 1 g/mL
Next, let's find the mass of glycerol in the solution in grams.
Mass of glycerol = 2m g - mass of water
= 2m g - m g
= m g
Now, we need to convert the mass of glycerol to moles.
To do this, we need the molar mass of glycerol.
The molar mass of C3H8O3 (glycerol) = (3*12.01 g/mol) + (8*1.01 g/mol) + (3*16.00 g/mol)
= 92.09 g/mol
Now, let's calculate the number of moles of glycerol:
Number of moles of glycerol = mass of glycerol / molar mass of glycerol
= m g / 92.09 g/mol
Finally, we can calculate the molality of glycerol in the solution.
Molality (m) = number of moles of solute / mass of solvent in kg
= (m g / 92.09 g/mol) / (m g / 1000 g/kg)
The 'm' on both sides cancels out, and our final equation becomes:
Molality (m) = g/mol / 10
Substituting the value of m = 1, we get:
Molality (m) = 1 / 10
= 0.1 m
Therefore, the molality of glycerol in the solution is 0.1 m (rounded to one decimal place).
So, it seems there was an error in your calculation. The correct answer is 0.1 m, not 11.9 m.