Question

Part A - Using the definition for mole fraction. A solution was prepared by dissolving 140.0 g of KCl in 305 g of water. Calculate the mole fraction of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.) Express the mole fraction of KCl to two decimal places. View Available Hint(s) χ χ = nothing Part B - Use the definition of molality A solution was prepared by dissolving 140.0 g of KCl in 305 g of water. Calculate the molality of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.) Express the concentration of KCl in molality to two decimal places. View Available Hint(s) Express the concentration of in molality to two decimal places. 0.16 mol/kg 0.01 mol/kg 6.15 mol/kg 0.10 mol/kg

Answer 1

Answer:

a) X KCl = 0.09

b) m KCl = 6.15 mol/Kg

Explanation:

a) mol fraction (X) = mol solute/mol total;  mol total = mol solute + mol solvent

⇒ X KCl = n KCl / n KCl + n H2O

∴ n KCl = 140.0g * ( mol / 74.6g ) = 1.876 mol KCl

∴ n H2O = 305g * ( mol / 18g ) = 16.944 mol H2O

⇒ X KCl = 1.876 / ( 1.876 + 16.944 ) = 0.09

b) molality (m) = mol solute / Kg solvent

∴ mol KCl = 140.0g KCl * ( mol / 74.6g ) = 1.876 mol KCl

∴ Kg H2O = 305g H2O * Kg/1000g = 0.305Kg H2O

⇒ m KCl = 1.876 / 0.305 = 6.15 mol/Kg