CHEM 1280 Lab Quiz 2 Definitions

Freezing point

the change from liquid to solid freezing and the temperature at which the transition occurs. This can be determined by temperature and particular intermolecular interactions.

Molality

the number of moles of solute in one kilogram of solvent or the freezing point depression (delta Tf) per Kc molal-1

Overall order of reaction

the sum of the orders with respect to each reactant

Colligative properties

properties of solutions that are affected by the number of solute particles present, regardless of their identity.

Define and calculate Molality (mc) using both mol of solute per mass of solvent, [kg] AND mc= ∆Tf/ Kf; be able to explain the differences in values of molality calculated from moles of solute and molality obtained from freezing point depression data for ionic substances

-Classic definition of molality: mol solute/kg solvent - The two molalities should be very close together - However sometimes the molalities are farther apart. When you are dealing with an ionic solute in a polar solvent, the molecules will completely separate. So, the molality found using freezing point depression would be greater than other molality because there is an increase in the number solute of particles. If there is a greater amount of particles it changes the magnitude of the freezing point depression. -Vant Hoff factor

Consider the following potential solutes (down below) and tell which is likely to dissolve in water and why, and which is unlikely to dissolve in water and cyclohexane (C6H12) and why: Solutions: Ethanol (C2H5OH), Potassium chloride (KCl), 1- Octene (C8H16)

1. Dissolve in water: Ethanol and Potassium Chloride (KCl). Both of these compounds have similar bonds and properties of water. - Ethanol: has hydrogen bonding and is polar. "Like dissolves like" and water has hydrogen bonding like ethanol so therefore, it dissolves in water - KCl: has ionic bonding and dissociates completely. Therefore, if it is put in a solvent that is polar like water, it will dissolve 2. Won't dissolve in water: 1-Ocetene (C8H16) is non-polar and it doesn't have similar bonding to water so it won't dissolve 3. Dissolve in Cyclohexane: 1-Octene (C8H16) is non-polar and so is cyclohexane so they have similar bonding properties and therefore it will dissolve in cyclohexane. - Ethanol can slightly dissolve in cyclohexane because it has a non-polar group C2H5 that is similar to cyclohexane's non-polar group. It is slightly soluble 4. Unlikely to dissolve in cyclohexane: potassium chloride because it is polar and therefore doesn't have similar properties to cyclohexane which is non-polar

Would the measured boiling point of your unknown liquid have been too high or too low if the thermometer bulb had rested on the bottom of the water-filled beaker? Explain

Boiling point would be too high cause the thermometer bulb would be measuring the temperature of glass which is directly under the bunsen burner flame and this would increase temperature.

Briefly explain why the procedure you used to determine boiling point would not work for liquids with boiling points greater than 100 degrees C.

Can't use this procedure with liquids greater than 100 cause water's boiling point is 100 C. The water bath would begin to evaporate so you couldn't find the boiling point of the liquid.

A student following the procedure described in this module used water as the solvent and encountered some interesting problems. Comment on the effect, if any, each of the following situations could have had on the experimental results. a. The unknown, a white powder, failed to dissolve in the solvent

If the powder doesn't dissolve you wouldn't get the right freezing point calculation, which would get the wrong freezing point and change molality. If these change your results are wrong.

A student determined the Kf of t-butyl alcohol using tap water instead of distilled or deionized water. Describe the problems that might have been encountered. How would these problems affect the magnitude of Kf?

Since tap water has ions in it, it affects the freezing point. Because of that, deionized and tap water don't have the same freezing points. This could throw off the Kf value because the freezing point helps determine Kf. Since tap water has a slightly higher freezing point, Kf could be a larger value.

Which substances would hydrogen bonds likely form and what effect did their formation would have on the boiling point of the substance and its solubility in water?

Substances that form hydrogen bonds are compounds with an OH group, an NH group, and an FH group. Hydrogen bonding affects boiling point by raising the boiling point of a compound that has hydrogen bonding present. Hydrogen bonding is a strong intermolecular force that raises the boiling point. Substances like ammonia and ethanol are soluble in water because they have hydrogen bonding which is also present in water. Like dissolves like

Compare the values of the normal boiling point of the following substances and explain the differences: Substances - Propane, C3H8; Boiling point: -42 - Propanone, CH3COCH3; Boiling point: 56 - n-propanol, C3H7OH; a. Boiling point: 100

The boiling point for n-propanol is the highest and propanone is significantly higher than propane with a boiling point at -42 degrees. The reason that n-propanol has a higher boiling point is that it has the strongest intermolecular force and the highest mass. N-propanol has hydrogen bonding which is the strongest intermolecular force, the stronger the intermolecular force, the higher the boiling point. Propane has the lowest boiling point because it has the weakest intermolecular force which is the London Dispersion force and the lowest mass. Propanone has dipole-dipole interactions which allow it to have a slightly high boiling point.

Order of reaction with respect to a certain reagent

The effect of changes in the concentration of just that reactant on the reaction rate

A student following the procedure described in this module used water as the solvent and encountered some interesting problems. Comment on the effect, if any, each of the following situations could have had on the experimental results. b. The student returned to the laboratory instructor for a different solid unknown. This unknown dissolved, but bubbles were seen escaping from the solution almost immediately after the addition of the solid.

The formation of bubbles indicates there was a reaction that took place. When gas is escaping, mass is lost and the number of solute particles is also decreased. This means that molality would decrease because there are fewer solute particles present and it would also decrease delta Tf.

Briefly explain why it is critical that the test tube containing the sample of nitrobenzene be dry when determining the freezing temperature of nitrobenzene

The test has to be dry because it would affect the freezing point if it was wet. If it changes the freezing point determination would mess up the calculations and would cause the freezing point to lower

A student following the procedure described in this module used water as the solvent and encountered some interesting problems. Comment on the effect, if any, each of the following situations could have had on the experimental results c. As the student was setting up the apparatus to measure the freezing point of the unknown solution, the thermometer assembly rolled off the laboratory bench, and the thermometer broke. The student obtained a new thermometer and experimented as instructed

This new thermometer could record different results than the first one because each tool is calibrated differently. This could throw off results

Given a cooling curve be able to find the freezing point of the pure solvent and a solution of an unknown. Determine freezing point depression.

a. Freezing point depression = Tf (pure solvent) - Tf' (freezing point of solution) b. Freezing point depression = Kf value * mc (molality in terms of mol solute/kg solvent)

If there is an experiment with the rate law rate= k[A]^2[B]^1, how will the rate change if: - Concentration of "A" was doubled - Concentration of "B" was cut in half

a. If A were doubled, then the rate would be quadrupled because the order with respect to A is squared meaning that the overall rate of the reaction would quadruple. b. If B was halved, then the reaction would be cut in half because the order of the reaction with respect to B is one and directly proportional to the reaction rate.

You weigh three samples of NaCl and dissolve each in 1.000 kg H2O. You then measure the freezing temperature of each solution and compare these temperatures to the freezing point of water. The data you collect are tabulated below. Explain the observed results. Predict and briefly explain the result you would expect for a solution made up of 29.22 g NaCl dissolved in 1.000 kg H2O

a. You have to find the molality of each gram NaCl, then the freezing point, then the ratio b. The closer the ratio is to two the better because the NaCl dissociates into 2 molecules. c. When the concentration is lower, the ratio is closer to 2. So, if 29.22g NaCl is dissolved into 1.0 kg H2O the ratio will be farther away from 2, so it won't be as dissociated

Normal boiling point

it is a temperature at which vapor pressure above liquid is equal to normal atmospheric pressure