Answer to Question #302111 in General Chemistry for Hutchii

Question

Calculate the molarity, molality, and mole fraction of NH3 for a solution of 30.0 g NH3 in 70.0 g H2O. The density of the solution is 0.982 g/mL.

Answer 1

Molarity = moles of solute / volume of solution = n / V

n(NH₃) = m(NH₃) / Mr(NH₃) = 30.0 g / 17.02 g/mol = 1.76 mol

V(solution) = m(solution) / $\rho$ (solution) = (30.0 g NH₃ +O 70.0 g H₂O) / 0.982 g/mL = 101.83 mL = 0.1018 L

M = 1.76 mol / 0.1018 L = 17.3 mol/L

Molality = moles of solute / kg of solvent

m = 1.76 mol NH₃ / 0.070 kg H₂O = 25.1 mol/kg

Mole fraction = moles of solute / moles of solution

n(H₂O) = m(H₂O) / Mr(H₂O) = 70.0 g / 18.015 g/mol = 3.88565 mol

X(NH₃) = 1.76 mol NH₃ / (1.76 mol NH₃ + 3.88565 mol H₂O) = 0.3117