Intermolecular Forces of Attraction - UIC Department of Chemistry

Chemistry 112 Worksheet Week 9 Fetzer GislasonA. Intermolecular Forces of Attraction1.Classify each of the following substances as ionic, polar, or nonpolar in nature.a) KBr b) O 2 c) Al 2 (Cr 2 O 7 ) 3 d) HCl e) Ar f) H 2 O2.Name the attractive forces between the following. Choose from : ion-dipole, dipole-dipole,dipole-induced dipole, induced dipole-induced dipole (dispersion forces).a) O 2 and HClb) KBr and HClc) H 2 O and HCld) Ar and O 2e) Ar and H 2 OB. Colligative Properties and molar massThe colligative properties of a solution are properties that depend only on the number of particlesdissolved in the solution, rather than on the kind (ions or molecules) of particles dissolved.How many particles will there be of each of the following units in water solution?a) C 2 H 5 OH b) Al 2 (Cr 2 O 7 ) 3 c) MgCl 2There are four colligative properties of solutions: vapor pressure lowering, boiling point elevation,freezing point depression, and osmotic pressure. The experiment this week studies freezing pointdepression. Molecules of a liquid have random chaotic motion and are held together by attractiveforces between the molecules. As the temperature of the liquid is decreased, the average kineticenergy of the molecules decrease (kinetic theory) and collisions become less vigorous. At thefreezing point, the attractive forces overcome the liquid’s random motion; a solid forms.If the liquid contains a solute, the solute molecules interfere with the organizing of the liquidmolecules into a solid. Then the kinetic energy must be reduced further before freezing takesplace. Thus the freezing point of the solution is lower than the freezing point of the pure liquid(solvent). For all dilute solutions,∆T = −K f m, where ∆T = T f (solution) – T f (pure solvent), m is a concentration unit called molality,and K f is a constant for a particular solvent (freezing point depression constant).Molality ismoles solute. This is useful in finding molar mass as moles solute =kg solventgramsmolar mass .1) A pure solvent has a freezing point of −3.50°C. When a small amount of solute is added to it,the freezing point is −3.95°C. What is ∆T f in this case?2) Calculate molality for the following:a) 0.30 mol ethylacetate is added to 250.0 g cyclohexane.b) 5.00 grams glucose (C 6 H 12 O 6 ; molar mass = 180.2 g) is added to 500.0 g water.

Chemistry 112 Worksheet Week 9 Fetzer Gislasonc) 0.20 mL of ethylacetate (d = 0.7785 g/mL ) is added to 25.0 g of cyclohexane.Ethylacetate is CH 3 CH 2 OCOCH 3 .3) Camphor is a white solid that melts at 179.5°C. K f for camphor is 40°C/m. A 1.07 g sample of acompound was dissolved in 78.1 g of camphor. The solution melted at 176°C. What is the molar massof the compound?[1.6×10 2 g/mol]Note: In lab you will find the freezing point of a pure solvent, the freezing point of the solvent + aknown solute, and the freezing point of the solvent + an unknown solute. For both the known and theunknown solute experiment, you will prepare graphs of −∆T versus mL solute added. You will need tofind the slope of the lines that result. Be sure to include the point (0,0) in your graphs. You will findthe molar mass of the unknown solute from:molar mass (unknown)molar mass (known)slope (known) density (unknown)=slope (unknown) density (known)This formula is derived from the freezing point formula: ∆T = −K f m, and assumes that the volume ofpure solvent is the same for the experiments with the known solute and with the unknown solute.