Answer to Question #175989 in General Chemistry for Karley Eilts

Question

What is the freezing point depression of an aqueous solution of 0.75 m NaCl?

Answer 1

∆T = iKfm

kf is molal freezing point depression constant of solvent (1.86°C/m for water)

Mass of water = molar Mass of water = 18.01528g/mol × 1mol = 18.01528 g

= 18.01528g H2O×1Kg H2O/1000g H2O = 0.01801528 Kg H2O

Molality = moles of NaCl/Kg of water = 0.75/018 = 41.67 mol/Kg

NaCl -> Na+ + Cl-

1 m of NaCl gives 2m of dissolved particles : 1 m of Na+ ions and 1 mol of Cl- ions therefore NaCl i = 2

∆Tf = iKfm = 2 × 1.86°C kg/mol × 41.67 mol/Kg

= 155.0124 °C