collegative properties

Which mechanisms can explain how dissolving a solute into a pure liquid affects its vapor pressure?

1) Solvent molecules adhere to solute particles, which makes them less likely to evaporate from solution, lowering the vapor pressure. 2) Solute particles reduce the surface area available for solvent evaporation. They physically prevent the evaporation of solvent molecules. This reduces the vapor pressure above the liquid.

NaCl Van Hoff factor value

2

Suppose that 0.3 moles of potassium iodide are added to 800 mL of formic acid and a change in the boiling point of formic acid is observed. If dissolved in the same volume of formic acid, which of the following would cause an equivalent change in its boiling point? Select all that apply. A) 0.2 moles of calcium chloride B) 0.3 moles of sodium hydroxide C) 0.3 moles of glucose D) 0.6 moles of fructose

A,B,D

If 0.2 moles of sodium chloride are added to 1.5 moles of water, and the vapor pressure of pure water is 23.8 mmHg, approximately what is the new vapor pressure of the solution?

According to Raoult's law, the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent. First, we need to find the mole fraction of the solvent, which is equal to the # of moles of solvent divided by the total moles present in the solution (# moles solvent + # moles solute): 1.5/(1.5 + 0.2) = 0.8 Now that we know the mole fraction of water is 0.8, we can multiply this by the original vapor pressure to yield the new one: (0.8)(23.8 mmHg) = 19 mmHg.

Which of the following are ways to decrease the amount of time that it takes to cook pasta in a pot of water on a stovetop? Select all that apply. Assume that pasta is only added to the pot once boiling occurs. A) Increase the heat under the pot B) Add salt C) Cook the pasta at high altitudes D) Use less pasta

B Adding a large quantity of salt to a pot of water will increase the water's boiling point, allowing it to achieve a hotter temperature before it begins to boil. If the pasta is added to the pot at this point, it will cook more quickly.

Which of the following actions will decrease the freezing point of a flask of aqueous n-propanol by the greatest amount? A) Dissolving 0.2 moles of fructose in the flask B) Dissolving 0.6 moles of sucrose in the flask C) Dissolving 0.3 moles of calcium chloride in the flask D) Dissolving 0.4 moles of sodium chloride in the flask

C. One molecule of CaCl2 = 1 molecule of calcium + 2 molecules of chloride. Therefore, 0.3 moles of CaCl2 gives rise to 0.9 moles of solute molecules, which results in a large freezing point depression.

True or false: Adding a solute to a pure liquid disrupts interactions between solute molecules, raising the vapor pressure of the solution.

False. Adding a solute to a pure liquid makes solvent molecules more likely to remain liquid because they are attracted to solute particles. Dissolved solute molecules also reduce the surface area available for solvent evaporation. Therefore, solvent molecules are less likely to escape the solution in the presence of dissolved solute, which means that the liquid's vapor pressure actually decreases.

What is the boiling point of 500 mL of water after 540 g of glucose is added? (Note: The Kb of H2O is 0.5 K mol-1 kg and the molar mass of glucose is 180 g/mol.)

First, calculate the number of moles of glucose added to the water:540 g glucose x (1 mol glucose / 180 g glucose) = 3 moles glucose Next, note that the density of water is 1 kg/L, so 500 mL of water, or 0.5 L, is equal to 0.5 kg. Then, simply plug values into the boiling point elevation formula, noting that the Van't Hoff factor for glucose = 1: ΔT = 1 x (0.5 K mol-1 kg) x (3 mol glucose / 0.5 kg H2O) = 3 K Recall that the boiling point of pure water is 100 °C. Since adding solute increases the boiling point, the new boiling point will be 100 °C + 3 °C = 103 °C.

What happens to the freezing point of 2 L of phenol when 1 mol of calcium chloride is added at room temperature? (Note: The density of phenol is 1.07 g/mL at 20 °C and Kf = -7.3 mol-1 kg.)

First, we need the formula for freezing point depression, which is ΔT = iKfm. Next, we need to find the molality of the solution. Since the density of phenol is extremely close to 1, we can approximate the volume as the mass, meaning that 2 L of phenol is equal to about 2 kg of phenol. We can then calculate molality:Molality = (1 mol CaCl2 / 2 kg phenol) = 0.5 mol/kg We also need the Van't Hoff factor (i) for calcium chloride, which is 3 because each molecule of CaCl2 gives rise to three ions in solution. Now we're ready to plug values into the formula and solve for ΔT:ΔT = 3 x (-7.3 mol-1 kg) x (0.5 mol/kg) = 10.95 °C Since freezing point is a colligative property, it will decrease by 10.95 °C after we add 1 mol of calcium chloride.

What happens to the boiling point of water when the number of solute particles dissolved in it increases? Why?

It increases because the solute particles decrease the vapor pressure, thus increasing the temperature at which the vapor pressure is equal to atmospheric pressure

collegative properties

Properties that do not depend on the identity of the solute. Rather, they depend on the total number of dissolved particles in solution

According to Raoult's law:

Psolution = Xsolvent*Psolvent where Xsolvent is equal to moles of solvent divided by the total number of moles (# moles solvent + # moles solute)

How many moles of sucrose must be added to 6000 mL of pure water to lower its freezing point from 0 °C to -3 °C? (Note: Kf for H2O = -1.9 mol-1 kg.)

The formula for freezing point depression is ΔT = iKfm. Since the density of pure water is 1 kg/L, we can calculate the molality as:Molality = x mol sucrose/ 6 kg H2O Additionally, i = 1 for this solution, since sucrose does not dissociate into multiple solute particles in water. Plugging in these values and the -3 °C for ΔT gives us:-3 °C = 1 x (-1.9 mol-1 kg) x (mol sucrose / 6 kg H2O) Rearranging the equation to solve for moles of sucrose yields 9 mol.

How do the molality and molarity of a substance dissolved in water differ?

They do not differ. The molarity would be equal to molality because the density of water is 1 kg/L

True or false: Colligative properties only depend on the ratio of the number of solute molecules to the number of solvent molecules.

True

True or false: Osmotic pressure is a colligative property that is defined as the amount of pressure required to prevent the flow of solvent through a semipermeable membrane.

True

Suppose that 116 g of NaCl are dissolved in 2 L of ethanol at 20° C. What is the boiling point of this solution? (Note: The density of ethanol at 20 °C is 0.79 g/mL, pure ethanol boils at 79 °C, and the Kb of ethanol is 1.2 K mol-1 kg.)

We will need to use the formula for boiling point elevation, which is ΔT = iKbm, where m is the molality of the solution. Molality = mol NaCl / kg ethanol To get the moles of NaCl, we divide the mass (116 g) by the molar mass (58 g/mol), which yields 2 mol NaCl. Then, we need to use the density of ethanol to convert liters to kg: 0.79 g EtOH x (1000 mL / 1 L) x (2 L) x (1 kg / 1000 g) = 1.6 kg EtOH This yields: Molality = 2 mol NaCl / 1.6 kg EtOH = 1.25 mol/kg Substitute this value into the formula, along with i = 2 (the Van't Hoff factor for NaCl), and Kb = 1.2: ΔT = 2 (1.2 K mol-1 kg) (1.25 mol/kg) = 3 K Since adding solutes to a solvent always increases its boiling point, the boiling point must increase by 3 K or 3 °C, which makes 82 °C the correct answer.

The vapor pressure of a pure solvent is best reduced by adding:

a large amount of a nonvolatile solute to the solvent

With increasing elevation, Patm ________

decreases

The freezing point of 1 L of water could be decreased by:

dissolving a salt

boiling point is the temperature at which the vapor pressure is

equal to ambient atmospheric pressure

adding solutes to a solvent always ______ its boiling point

increases

Osmotic pressure increases with

increasing temperature

At higher elevations, solvents have __________ boiling points

lower

At high altitudes, the boiling point of water is

lower, because atmospheric pressure decreases, so water boils at a lower temperature

The vapor pressure of a solution, according to Raoult's Law, is equal to

mole fraction of the SOLVENT * the vapor pressure of the pure solvent As the mole fraction of the solvent decreases with the addition of solute, the solution's vapor pressure decreases proportionally

osmotic pressure

pressure that must be applied to prevent osmosis (solvent movement) across a selectively permeable membrane

the four basic collegative properties

vapor pressure reduction boiling point elevation freezing point depression osmotic pressure

boiling point elevation equation

ΔTb=iKbm where Kb is the boiling point constant m is the molality of the solute (m = mol solute/kg of solvent) i = Van Hoff factor, which refers to the dissolved particles generated when a substances ionizes