If you dissolve 220.3 Pounds of ammonium nitrate in water to obtain a final solution volume of 500. gallons what is the [NH4+] (of course in mol/L)?
Convert pounds to grams = ?
Convert grams to mols with mols = grams/molar mass = ?
Convert gallons to liters = ?
Then mols/L = [NH4^+] = ?
220.3 lbs = 100016.2 g
500. gallons= 1892.5L
100016.2g (NH4)(NO3)X 1 MOL/80.052 g/MOL= 1249.39 MOLES.
1249.39 MOLES/ 1892.5L = 6.60X10^-1 MOL/L (NH4)(NO3)
To calculate the molarity of ammonium nitrate ([NH4+]) in the final solution, you first need to convert the weight of ammonium nitrate from pounds to grams, and then determine the number of moles.
1. Convert pounds to grams: There are 453.592 grams in one pound. So, 220.3 pounds of ammonium nitrate is equal to 220.3 * 453.592 = 99,898.9976 grams.
2. Calculate the molar mass of ammonium nitrate (NH4NO3):
The molar mass of nitrogen (N) is 14.007 grams/mol.
The molar mass of hydrogen (H) is 1.008 grams/mol.
The molar mass of oxygen (O) is 16.00 grams/mol.
NH4NO3 = (1 * molar mass of N) + (4 * molar mass of H) + (3 * molar mass of O)
= (1 * 14.007) + (4 * 1.008) + (3 * 16.00)
= 14.007 + 4.032 + 48.00
= 66.039 grams/mol
3. Calculate the number of moles of ammonium nitrate in the solution:
Number of moles = mass of substance / molar mass
Number of moles = 99,898.9976 grams / 66.039 grams/mol
= 1,513.06 moles
4. Calculate the volume of the solution in liters:
As the given volume is in gallons, you need to convert it to liters.
1 gallon is approximately equal to 3.78541 liters.
Volume of the solution = 500 gallons * 3.78541 liters/gallon
= 1,892.705 liters
5. Finally, calculate the molarity ([NH4+]) of ammonium nitrate:
Molarity (M) = moles of solute / volume of solution in liters
Molarity ([NH4+]) = 1,513.06 moles / 1,892.705 liters
≈ 0.7989 mol/L
Therefore, the molarity of ammonium nitrate ([NH4+]) in the final solution is approximately 0.7989 mol/L.