A nonvolatile solute will not contribute to the vapor pressure of the solution, but will decrease the vapor pressure of the pure solvent. We can solve for the vapor pressure of the solution by using Raoult's law:

In other words, the new vapor pressure is equal to the molar fraction of the solvent multiplied by the vapor pressure of the pure solvent.

Since both of the solvents have a vapor pressure, we can find the percentage of ethanol in the solution by using Raoult's law, including both solvents in the equation:

Since we want to find the percentage of ethanol in the solution, we will designate the molar fraction of ethanol as . Since the sum of the molar fractions equals one, the molar fraction of hexane will be designated as .

So, 63.4% of the solution is ethanol.

Since the solute is volatile, it will contribute to the total vapor pressure of the solution. As a result, we must incorporate it when solving for the total vapor pressure.

We can find the vapor pressures using Raoult's law.

Since we are solving for the molar fraction of the solvent, we will designate its molar fraction as . The sum of the molar fractions of each component must be equal to 1; thus, the molar fraction of the solute must be . Using these variables and the information given, we can solve for the molar fraction of the solvent.

The equation for freezing point depression is , where  is the change in temperature,  is a constant related to the solvent,  is molality, and  is the van't Hoff factor, which is the number of ion particles from each dissolved molecule. We simply plug these numbers into the equation to find the new freezing point.

We know our constant is . Molality is moles of solute per kilogram of solution, and we know that the density of water is one kilogram per liter and the molecular weight of sodium chloride is .

The van't Hoff factor is . Sodium chloride creates only two ions when dissolved, one  and one .

Using these values, we can solve for the freezing point depression.

The freezing point will be decreased by . The normal freezing point is , making the new freezing point .

Since a salt has been added to the pure water, we can find the new melting point of the solution by using the freezing point depression equation:

The change in temperature is equal to the freezing point constant for the solvent multiplied by the molality of the solution multiplied by the van't Hoff factor. The van't Hoff factor is the number of ions that a salt will dissociate into when in solution.

For this particular salt, the van't Hoff factor is three. The molality will be equal to the moles of solute over the mass of the solvent.

Using these terms together in the original equation, we can find the freezing point depression.

This is the change in temperature from the regular freezing point. Since the freezing point for pure water is , the new melting point is .

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L = 

A simple calculation can be done to perform any solution dilution problem. We know our equation .

We can rewrite this as .

Using this formula, we take the old solution and set it equal to the new solution.

We need 0.16 liters of our 10 molar solution.

This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.

Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.