Question

What is the A) molality and B) mole fraction of 825 mg of Na3PO4 dissolved in 450mL of water? (Molar Mass (MM) of Na3PO4 = 163.94 g mol-1)

Answer 1

The molality of Na3PO4 is 0.0112 mol/kg, and the mole fraction is 0.000201 in the given solution.

Step-by-step explanation:

Calculating Molality and Mole Fraction

To find the molality of Na3PO4 in the solution, you need to know the number of moles of solute and the mass of the solvent in kilograms. First, convert the mass of Na3PO4 from milligrams to grams (825 mg = 0.825 g). Then, use the molar mass (163.94 g/mol) to find the number of moles:

moles of Na3PO4 = 0.825 g ÷ 163.94 g/mol = 0.00503 mol

Convert the mass of water from milliliters to kilograms (450 mL = 0.45 kg) and calculate the molality:

molality (m) = moles of solute ÷ mass of solvent (kg)

molality = 0.00503 mol ÷ 0.45 kg = 0.0112 mol/kg

To find the mole fraction of Na3PO4, calculate the moles of water (H2O) using its density (1 g/mL) and molar mass (18.015 g/mol):

moles of H2O = 450 g ÷ 18.015 g/mol = 24.98 mol

Then, use the mole fraction formula:

mole fraction of Na3PO4 = moles of Na3PO4 ÷ (moles of Na3PO4 + moles of H2O)

mole fraction = 0.00503 ÷ (0.00503 + 24.98) = 0.000201