How many grams of ethylene glycol, HOCH2CH2OH, are present in 5.3 gallons of a 5.22 m solution with a density of 1.83 g/mL?
I would do this.
5.22m means 5.22 moles in 1 kg solvent. If we take 1 kg solvent we will have 5.22 x molar mass ethylene glycol = 5.22*62 = about 324 g Ethgly. The solution will have a mass of 1000 g + 324 = about 1324 (You need to go through and redo since I've estimated here and there.). So how many grams ethylene glycol must we have in a 5.3 gallon. I typed 5.3 gallons to liters into google and it returned 20.063 L or 20063 mL and that x 1.83 gives a mass of about 36,716 grams for the solution.
So moles ethylene glycol = 324 x (36715/1324) = = about 8985 g ethylene glycol.
I like to check these things to see if it will make a 5.22 m soln.
8984/62 = 145 moles ethylene glycol.
mass soln = 36715 and subtract 8984 g for the solute = about 27,700 g or 27.7 kg. Then 145 moles/27.7 kg = 5.23m which is a good answer considering I rounded here and there.
To calculate the number of grams of ethylene glycol (HOCH2CH2OH) present in 5.3 gallons of a 5.22 m (molar) solution with a density of 1.83 g/mL, we need to follow these steps:
Step 1: Convert gallons to milliliters.
To convert gallons to milliliters, we can use the conversion factor: 1 gallon = 3,785.41 milliliters.
Therefore, 5.3 gallons is equal to 5.3 x 3,785.41 = 20,075.773 milliliters.
Step 2: Calculate the volume of the solution in milliliters.
The solution's volume is given as 20,075.773 milliliters.
Step 3: Calculate the mass of the solution in grams.
The density of the solution is given as 1.83 g/mL.
Using the formula: mass = density x volume, we can calculate the mass of the solution:
mass = 1.83 g/mL x 20,075.773 mL = 36,738.131 g.
Step 4: Calculate the moles of ethylene glycol.
The molarity (m) of the solution is given as 5.22 m (molar).
This means the solution contains 5.22 moles of ethylene glycol per liter.
To find the moles of ethylene glycol, we need to convert the solution's volume from milliliters to liters: 20,075.773 mL ÷ 1000 = 20.075773 liters.
moles of ethylene glycol = molarity x volume in liters
moles of ethylene glycol = 5.22 m x 20.075773 L = 104.103961173 moles.
Step 5: Convert moles to grams.
Using the molar mass of ethylene glycol (HOCH2CH2OH), which can be calculated from the atomic masses:
C = 12.01 g/mol (atomic mass of carbon)
O = 16.00 g/mol (atomic mass of oxygen)
H = 1.01 g/mol (atomic mass of hydrogen)
The molar mass of ethylene glycol (HOCH2CH2OH) is: 2(12.01) + 2(1.01) + 2(16.00) = 62.07 g/mol.
To convert moles to grams, we multiply the moles of ethylene glycol by its molar mass:
grams of ethylene glycol = moles of ethylene glycol x molar mass
grams of ethylene glycol = 104.103961173 moles x 62.07 g/mol = 6,459.63325412 g.
Therefore, there are approximately 6,459 grams of ethylene glycol (HOCH2CH2OH) present in 5.3 gallons of a 5.22 m solution with a density of 1.83 g/mL.