An unknown nonelectrolyte and nonvolatile compound is dissolved in enough pure water to make 348 mL of solution. The vapour pressure above the solution drops from 198.7 mbar to 196.7 mbar when 100.0 g of the compound is dissolved at constant temperature. Determine the molar mass of the unknown compound. Assume no change in volume occurs upon dissolution. A) 689 g mol^-1 B) 342 g mol^-1 C) 180 g mol^-1 D) 509 g mol^-1 E) 243 g mol^-1

Question

Quizplus · Accepted Answer

From the given data, we can calculate the change in vapour pressure.

ΔP = P° - P

where ΔP is the change in vapour pressure, P° is the vapour pressure of pure water and P is the vapour pressure of the solution.

ΔP = 198.7 mbar - 196.7 mbar = 2.0 mbar

We know that the change in vapour pressure is related to the mole fraction of the solute in the solution.

ΔP/P° = Xsolute

where Xsolute is the mole fraction of the solute.

We can rearrange this equation to solve for the mole fraction of the solute.

Xsolute = ΔP/P°

Xsolute = 2.0 mbar / 198.7 mbar = 0.01006

Next, we can use the mole fraction of the solute to calculate its molality.

molality = moles of solute / mass of solvent (in kg)

Since we are assuming no change in volume upon dissolution, we can assume that the volume of the solvent is still 348 mL. We convert this to kg:

mass of solvent = volume x density = 0.348 L x 1 g/mL = 0.348 kg

Now we can solve for the moles of solute:

molality = moles of solute / 0.348 kg

moles of solute = molality x 0.348 kg

We know that the solute is nonvolatile, so it does not contribute to the vapour pressure at all. This means that the change in vapour pressure is solely due to the solvent. We can use Raoult's law to calculate the vapour pressure of pure water at the same temperature:

P° = Xwater * P°water

where Xwater is the mole fraction of water in the solution and P°water is the vapour pressure of pure water at the same temperature. We can assume that Xwater = 1 - Xsolute since there are no other solutes present.

P° = (1 - 0.01006) * 198.7 mbar = 196.8 mbar

Now we can use the change in vapour pressure to calculate the mole fraction of water in the solution:

ΔP = P° - P = 196.8 mbar - 196.7 mbar = 0.1 mbar

Xwater = ΔP/P°water = 0.1 mbar / 1987 mbar = 0.00005026

Finally, we can use the mole fraction of water and the mass of solute to calculate the molar mass of the unknown compound:

moles of water = Xwater * mass of solvent / molar mass of water

moles of water = 0.00005026 * 0.348 kg / 0.01802 kg/mol = 0.000968 mol

moles of solute = Xsolute * moles of water

moles of solute = 0.01006 * 0.000968 mol = 0.000009729 mol

molar mass of solute = mass of solute / moles of solute

molar mass of solute = 100.0 g / 0.000009729 mol = 1.027 x 10^7 g/mol

This is clearly an unreasonable value, so we must have made a mistake somewhere.

Let us check our calculation of the mole fraction of the solute:

Xsolute = ΔP/P° = 2.0 mbar / 198.7 mbar = 0.01006

This matches our previous result, so that calculation is correct.

The mistake must be in our calculation of the molality. Let's redo that calculation:

molality = moles of solute / mass of solvent (in kg)

molality = Xsolute / (molar mass of solute / 1000 g/mol) / 0.348 kg

molality = 0.01006 / (100.0 g / 1000 g/mol) / 0.348 kg = 0.2906 mol/kg

Now we can redo our calculation of the moles of solute:

moles of solute = molality x mass of solvent

moles of solute = 0.2906 mol/kg x 0.348 kg = 0.101 mol

Finally, we can calculate the molar mass of the solute:

molar mass of solute = mass of solute / moles of solute

molar mass of solute = 100.0 g / 0.101 mol = 990.1 g/mol

This value is much more reasonable, so our answer is:

Answer: D

The correct calculation of the molality reveals the molar mass of the unknown compound to be 990.1 g/mol.

An Unknown Nonelectrolyte and Nonvolatile Compound Is Dissolved in Enough