You wish to prepare an aqueous solution of glycerol, C3H5(OH)3, in which the mole fraction of the solute is 0.093. How many grams of glycerol must you combine with 425 grams of water to make this solution? What is the molality of the solution?
PLEASE HELP!
Do you know what the mole fraction is?
Xgly=(nly/(ng+ nH2O.
You know grams water, convert that to moles. You know moles glycerol = grams/molar mass and you know molar mass. You know Xglycerol = 0.093.
Solve for grams glycerol which is the only unknown.
After finding the moles glycerol, molality can be calculated from the definition of molality.
m = moles solute/kg solvent.
Post your work if you get stuck.
To determine the number of grams of glycerol needed to make the solution, we first need to find the mole fraction of glycerol in the solution.
The mole fraction is given by the equation:
Mole fraction of glycerol (xglycerol) = moles of glycerol / total moles
Since the mole fraction of glycerol is given as 0.093, we can write:
0.093 = moles of glycerol / (moles of glycerol + moles of water)
We know that the molecular weight of glycerol (C3H5(OH)3) is:
(3 × 12.01 g/mol) + (5 × 1.01 g/mol) + 3 × (16.00 g/mol) = 92.09 g/mol
Let's assume the number of moles of glycerol is nglycerol, and the number of moles of water is nwater.
Using the given information, we know that the number of grams of glycerol is equal to the number of moles of glycerol multiplied by the molecular weight of glycerol:
Grams of glycerol = nglycerol × (92.09 g/mol)
We also know that the grams of water is equal to the number of moles of water multiplied by the molecular weight of water (18.02 g/mol):
Grams of water = nwater × (18.02 g/mol)
Given that there are 425 grams of water, we can calculate the number of moles of water using the equation:
425 g = nwater × 18.02 g/mol
nwater = 425 g / 18.02 g/mol
Now, let's substitute these values into the mole fraction equation:
0.093 = nglycerol / (nglycerol + (425 g / 18.02 g/mol))
Rearranging this equation to solve for nglycerol, we have:
nglycerol = 0.093 × (nglycerol + (425 g / 18.02 g/mol))
nglycerol = 0.093 × (nglycerol + 23.603)
nglycerol = 0.093nglycerol + 2.19
0.907nglycerol = 2.19
nglycerol = 2.41 moles
Now we can substitute this value of nglycerol into the equation for grams of glycerol:
Grams of glycerol = 2.41 moles × (92.09 g/mol)
Grams of glycerol = 222.09 grams
Therefore, in order to prepare an aqueous solution in which the mole fraction of glycerol is 0.093, you must combine 222.09 grams of glycerol with 425 grams of water.
Now let's calculate the molality of the solution:
Molality (m) is defined as moles of solute per mass of solvent in kilograms. In this case, the solute is glycerol and the solvent is water.
Molality (m) = moles of glycerol / mass of water in kg
Moles of glycerol = 2.41 moles (as we calculated before)
Mass of water = 425 grams = 0.425 kg
Molality (m) = 2.41 moles / 0.425 kg
Molality (m) = 5.677 mol/kg (rounded)
Therefore, the molality of the solution is approximately 5.677 mol/kg.
To determine how many grams of glycerol you need to combine with 425 grams of water, you will first need to calculate the mass of the solution. Then, using the mole fraction, you can determine the mass of glycerol.
Step 1: Calculate the mass of the solution
Mass of solution = mass of water + mass of glycerol
Given:
Mass of water = 425 grams
Step 2: Calculate the mass of glycerol
Let's assume the mass of glycerol is x grams.
Mass of glycerol + Mass of water = Mass of solution
x grams + 425 grams = Mass of solution
Step 3: Calculate the mole fraction of glycerol
The mole fraction of glycerol (Xglycerol) is defined as the moles of glycerol divided by the total moles of the solution.
Xglycerol = moles of glycerol / (moles of water + moles of glycerol)
The mole fraction of glycerol is given as 0.093, so:
0.093 = moles of glycerol / (moles of water + moles of glycerol)
Since the mole fraction is defined as the ratio of moles, we can write:
0.093 = moles of glycerol / (moles of water / molar mass of water + moles of glycerol / molar mass of glycerol)
Step 4: Calculate the molar mass of water and glycerol
Molar mass of water = 18 g/mol
Molar mass of glycerol = 92 g/mol
Step 5: Calculate the moles of water and glycerol
moles of water = mass of water / molar mass of water
moles of glycerol = mass of glycerol / molar mass of glycerol
Substituting the values into the equation from step 3, we have:
0.093 = moles of glycerol / (moles of water / 18 + moles of glycerol / 92)
Step 6: Solve for moles of glycerol
Multiply both sides of the equation by the denominator to get:
0.093 * (moles of water / 18 + moles of glycerol / 92) = moles of glycerol
0.093 * (425 g / 18 + moles of glycerol / 92) = moles of glycerol
Step 7: Solve for moles of glycerol
Now we can solve for moles of glycerol:
0.093 * (425 / 18 + moles of glycerol / 92) = moles of glycerol
Multiply both sides of the equation by 92 to eliminate the denominator:
0.093 * (425 / 18 * 92) + 0.093 * moles of glycerol = moles of glycerol
0.093 * (425 * 5.111) + 0.093 * moles of glycerol = moles of glycerol
214.44 + 0.093 * moles of glycerol = moles of glycerol
Rearrange the equation to isolate the moles of glycerol:
0.093 * moles of glycerol - moles of glycerol = -214.44
(0.093 - 1) * moles of glycerol = -214.44
-0.907 * moles of glycerol = -214.44
moles of glycerol = -214.44 / -0.907
moles of glycerol = 236.501
Step 8: Calculate the mass of glycerol
mass of glycerol = moles of glycerol * molar mass of glycerol
mass of glycerol = 236.501 * 92
mass of glycerol ≈ 21707.692 grams (rounding to four decimal places)
Therefore, you need approximately 21707.692 grams of glycerol to combine with 425 grams of water to make the solution.
Now, let's calculate the molality of the solution.
Molality is defined as the moles of solute per kilogram of solvent. In this case, the solvent is water.
Step 1: Convert the mass of water to kilograms
Mass of water = 425 grams
Mass of water in kilograms = 425 grams / 1000 = 0.425 kg
Step 2: Calculate the molality of the solution
Molality (m) = moles of glycerol / mass of water in kilograms
Molality = moles of glycerol / 0.425 kg
Molality = 236.501 / 0.425
Molality ≈ 556.474 mol/kg (rounded to three decimal places)
Therefore, the molality of the solution is approximately 556.474 mol/kg.