Question

292.5 g of NaCl completely dissolves (producing Na+ and Cl– ions) in 1.00 kg of water at 25.0°C. The vapor pressure of pure water at this temperature is 23.8 torr. Determine the vapor pressure of the solution?

Answer 1

The vapor pressure of the solution formed by dissolving 292.5 g of NaCl in 1.00 kg of water at 25.0°C is calculated using Raoult's Law, and it is found to be 20.16 torr.

Step-by-step explanation:

The question asks to determine the vapor pressure of a solution formed when 292.5 g of NaCl is dissolved in 1.00 kg of water at 25.0°C, given the vapor pressure of pure water is 23.8 torr at this temperature. To find the vapor pressure of the solution, we can use Raoult's Law, which relates the vapor pressure of a solution to the mole fraction of the solvent and the vapor pressure of the pure solvent.

To use Raoult's Law, we need to:

1. Calculate the number of moles of NaCl.
2. Determine the mole fraction of water in the solution.
3. Apply Raoult's Law to find the vapor pressure of the solution.

First, find the moles of NaCl using its molar mass (58.44 g/mol):

(292.5 g NaCl) / (58.44 g/mol) = 5.00 moles NaCl

Since NaCl dissociates into Na+ and Cl−, the total moles of solute particles will be double:

5.00 moles NaCl × 2 = 10.00 moles of particles

Now, calculate the moles of H2O using its molar mass (18.02 g/mol):

(1000 g H2O) / (18.02 g/mol) = 55.49 moles H2O

The mole fraction of water is then:

(55.49 moles H2O) / (55.49 moles H2O + 10.00 moles solute particles) = 0.847

Finally, apply Raoult's Law to find the vapor pressure of the solution (P_solution):

P_solution = (mole fraction of water) × (vapor pressure of pure water)

P_solution = 0.847 × 23.8 torr = 20.16 torr

Therefore, the vapor pressure of the solution is 20.16 torr.

Answer 2

21.8 torr

Step-by-step explanation:

Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this

Psol= Xsolvent * P°solvent where,

Psol - the vapor pressure of the solution

Xsolvent - the mole fraction of the solvent

P°solvent - the vapor pressure of the pure solvent

So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.

To do that, use the two compounds' respective molar masses. For sodium chloride you'll have

292.5g * (1mol NaCl/58.44g) = 5.005 moles of NaCl

Same for water except you first have to convert to grams by multiplying by 1000

1000g * (1 mol H2O/18.015g) = 55.51 moles H2O

Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.

ntotal = nwater +nNaCl

ntotal = 55.51 + 5.005 = 60.515 moles

The mole faction of water will thus be

Xwater= 55.51 moles/60.515 moles = 0.917

This means that the vapor pressure of the solution will be

Psol= 0.917 * 23.8 torr = 21.8 torr