ll, colligative properties of solutions (22 points), when 20.0 grams of an unkno

Question

ll, colligative properties of solutions (22 points), when 20.0 grams of an unknown compound are dissolved in 500. grams of benzene (CEHE), the freezing point of the resulting solution is 3.77 c. The freezing point of pure benzene c, and the K benzene 5.12.cym. The density of the solution is glmL For this solution, calculate the following: A. Molality of the solution (4 points) B. Molar mass of the unknown (6 points) Molarity of the solution (6 points) D. Mole Fraction of unknown in the solution. (6 points)

Explanation / Answer

1. molality =deltaT/Kf=1.67/5.12=0.326m

moles = 0.326X.500kg=0.163

2. moles= mass of solute/MM; MM= mass of solute/moles=20/0.163=122.7g/mol; Molar mass=122.7g/mol

3.Molarity =moles/L of solution; 0.879=mass of solution/L; M=0.326mol/0.526L=0.551M

4.mole fraction = moles of solute/moles of solution= 0.326/0.332=0.984

moles of solution = moles of solvent+moles of solute

moles of solvent= 500/78= 6.41; moles of solute =20/122.7=0.163 ; mole fraction = 0.0248

mole fraction = moles of solute/total molesof solution

0,163/6.573= 0.0248

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