Question

Assume a 3.0 mass % aqueous solution of compound that has molar mass of 74.44. (Assume the density of water at this temperatue is 1.00 g/mL.) Part A What is the molality of the compound in this solution? Express your answer using two significant figures. IVO AQ R O 2 ? molality = Submit Request Answer Part B What is the mole fraction of this comound in the solution? Express your answer using two significant figures. VA¢ * N O 2 ? Submit Request Answer

Answer 1

Part-A of Givenas Mass 080 = 3.00). Molar mass= 74.44 g/mol 3.0 mass of of solution, which means each 100g solution contains 3 g of solute weight of solute = 39 weight of solvent (water) = 97 g we know that, Molality= no. of mole of solute weight of solvent (kg) weight of solute Molar mass of solute weight of solvent (kg) 39 74.44 g/mol = 0.4155 mol 97 kg Kg 1000 g

molality= 0.4155 mol/kg molality = 0.42 m Ang Part-B W mole fraction of compound = no. of mole of compound no. of mole of compoundt no. of mole of water 3 g 74.44 g/mol 3 g + 97 g 74.44 /mol 18 g/mol = 0.0403 0.0403+ 5.388 = 0.0403 5.4283 = 0·0074 mol fraction of compound Ans