How many grams of ethylene glycol, HOCH2CH2OH, are present in 5.6 gallons of a 4.81 m solution with a density of 1.98 g/mL?
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To find the number of grams of ethylene glycol present in 5.6 gallons of a 4.81 molar solution, we can use the following steps:
1. Convert gallons to milliliters: Since density is given in grams per milliliter, we need to convert gallons to milliliters by using the conversion factor: 1 gallon = 3,785.41 milliliters.
Calculate: 5.6 gallons * 3785.41 mL/gallon = 21303.6 mL
2. Calculate the number of moles of ethylene glycol:
We are given that the solution is 4.81 molar, meaning there are 4.81 moles of ethylene glycol per liter of solution.
Convert the mL to liters: 21303.6 mL / 1000 mL/L = 21.3036 L
Calculate the number of moles: 21.3036 L * 4.81 mol/L = 102.363 moles
3. Calculate the mass of ethylene glycol:
The molar mass of ethylene glycol (HOCH2CH2OH) can be found by summing the atomic masses of all the elements in the formula.
Carbon (C) atomic mass = 12.01 g/mol
Hydrogen (H) atomic mass = 1.01 g/mol (x 4 atoms)
Oxygen (O) atomic mass = 16.00 g/mol (x 2 atoms)
Calculate: (12.01 g/mol * 2) + (1.01 g/mol * 4) + (16.00 g/mol * 2) = 62.07 g/mol
Calculate the mass of ethylene glycol: 102.363 moles * 62.07 g/mol = 6,354.86 g
Therefore, there are approximately 6,354.86 grams of ethylene glycol present in 5.6 gallons of a 4.81 m solution with a density of 1.98 g/mL.