How many grams of Ammonium Nitrate would you use to make a 500 ml of 5.684 M solution?
Answer
To find the number of grams of Ammonium Nitrate needed to make a 500 ml solution of 5.684 M, we can use the formula for Molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
Given that the desired Molarity is 5.684 M and the volume is 500 ml (0.5 L), we need to calculate the moles of Ammonium Nitrate required.
Rearranging the formula, we get:
moles of solute = Molarity * volume of solution (in liters)
moles of solute = 5.684 M * 0.5 L
Now, to convert moles to grams, we need the molar mass of Ammonium Nitrate (NH4NO3). The molar mass of NH4NO3 is calculated by adding up the atomic masses of each element:
N (14.01 g/mol) + H (1.01 g/mol) * 4 + O (16.00 g/mol) * 3 = 80.05 g/mol.
Now, we can calculate the grams of Ammonium Nitrate required:
grams of Ammonium Nitrate = moles of solute * molar mass of NH4NO3
grams of Ammonium Nitrate = (5.684 M * 0.5 L) * 80.05 g/mol
Simply multiply:
grams of Ammonium Nitrate = 1421 g.
Therefore, you would need 1421 grams of Ammonium Nitrate to make a 500 ml 5.684 M solution.