State how you would prepare each solution:
1) 455g of a solution of C3H5(OH)3 in water in which the mole fraction of glycine is .250
2) 755g of 1.4 molality solution of napathalene C10H8 in benzene C6H6.
My instructor told us to find the amount of grams in order to get the given grams, which confused me, and I really need help and thank you.
What you have written is the formula for glycerine and not glycine.
Let G = grams glycerine
and 455-G = grams H2O
------------------------
n = mols; nG = mols glycerine; nH2O = mols H2O
molar mass glycerine = about 92
molar mass H2O = about 18.
Then [nG/(nG + nH2O)] = xglycerine = 0.25
nG = grams glycerine/92
nH2O = grams H2O = (455-G)/18
Substitute for nG and nH2O in the above equation and solve for grams glycerine which is the only unknown in the equation.
2. A 1.4 m solution of naphthalene in benzene will have 1.4 mols naphthalene in ? kg benzene.
molar mass naphthalene = 128
mols naphthalene = g N/128 which we will call N/128.
m = mols solute/kg solvent
kg benzene = 0.755 - kg N
1.4 = (N/128)
1.4 = (N/128)/kg benzene
1.4 = (N/128)/(0.755 - 0.001N)
Solve for N and I get something like 115 but that isn't exact.
Then benzene = 0.755-0.115 = about 0.64
So 115 g naphthalene in 0.64 kg benzene will give you
(115/128)/0.64 = 1.4m
To prepare each solution, you need to know the molar mass of the solute and the properties of the solvent.
1) To prepare a solution of C3H5(OH)3 in water with a mole fraction of glycine equal to 0.250, you need to calculate the amount of glycine (in grams) needed and then dissolve it in water.
Here's how you can determine the amount of glycine needed:
- Determine the mole fraction of glycine means that the mole fraction of water will be (1 - 0.250) because it is a binary solution. So, the mole fraction of water is 0.750.
- From the mole fraction of water, you can determine the mole fraction of glycine using the fact that the mole fractions add up to 1. Set up an equation like this:
0.250 = (moles of glycine) / (moles of glycine + moles of water)
- Rearrange the equation and isolate the moles of glycine:
moles of glycine = 0.250 * (moles of glycine + moles of water)
moles of glycine = 0.250 * moles of water / (1 - 0.250)
- Convert the moles of glycine to grams using the molar mass of glycine (C3H5(OH)3). Multiply the moles by the molar mass, which can be found on the periodic table or calculated:
grams of glycine = moles of glycine * molar mass of glycine (g/mol)
Once you have determined the grams of glycine needed, you can dissolve them in the appropriate amount of water to prepare the 455g solution.
2) To prepare a 1.4 molality solution of naphthalene (C10H8) in benzene (C6H6), you need to calculate the amount of naphthalene and then dissolve it in benzene.
Here's how you can determine the amount of naphthalene needed:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent. Therefore, you need to calculate the moles of naphthalene needed given the molality and the mass of the solvent.
- Determine the mass of benzene by subtracting the mass of the solution (755g) with the mass of naphthalene, which you're trying to find.
mass of benzene = mass of solution - mass of naphthalene
- Convert the mass of benzene from grams to kilograms, as molality is defined in terms of kg of solvent.
mass of benzene (kg) = mass of benzene (g) / 1000
Now, you can use the molality equation to find the moles of naphthalene needed:
molality = moles of naphthalene / mass of benzene (kg)
Rearrange the equation to isolate the moles of naphthalene:
moles of naphthalene = molality * mass of benzene (kg)
Once you have determined the moles of naphthalene needed, you can convert it to grams using the molar mass of naphthalene (C10H8) and dissolve it in the appropriate amount of benzene to prepare the 755g solution.
It's important to note that the molar masses of the solutes (C3H5(OH)3 and C10H8) and the solvents (water and benzene) should be known or determined in order to perform the calculations accurately. Make sure to use the correct molar masses for accurate results.