A saturated solution of ammonium sulfate, (NH4)2SO4, in water at 30 0C contains 78.0 g (NH4)2SO4 per 100.0 g H2O. What is the molality of this solution?
Molality = #mols/kg solvent
mols = g/molar mass
mols (NH4)2SO4 = 78.0g x (1 mol/molar mass) = ??
100.0 g H2O = 0.100 kg.
Plug and chug.
Post your work if you get stuck.
To find the molality of a solution, you need to determine the moles of solute and the mass of the solvent.
Step 1: Calculate the moles of solute (ammonium sulfate).
To find the moles of (NH4)2SO4, you can use its molar mass.
Molar mass of (NH4)2SO4 = (2 * 14.01 g/mol) + (4 * 1.01 g/mol) + 32.07 g/mol + (4 * 16.00 g/mol) = 132.14 g/mol
To convert grams to moles, use the formula:
moles = mass (g) / molar mass (g/mol)
moles of (NH4)2SO4 = 78.0 g / 132.14 g/mol
Step 2: Calculate the mass of the solvent (water).
The mass of water is given as 100.0 grams.
Step 3: Calculate the molality.
Molality is defined as the number of moles of solute per kilogram of solvent. Since mass of water is given in grams, we need to convert it to kilograms.
molality = moles of solute / mass of solvent (in kg)
mass of solvent (in kg) = mass of water (g) / 1000
Substituting the values,
molality = (78.0 g / 132.14 g/mol) / (100.0 g / 1000)
molality = 0.590 mol / 0.100 kg
molality = 5.90 mol/kg
Therefore, the molality of the solution is 5.90 mol/kg.