Question

A solution contains 15.27 grams of NaCl in 0.670 kg water at 25 °C. What is the vapor pressure of the solution?

Answer 1

Answer:

The vapor pressure of the solution is 23.636 torr

Explanation:

$P_{solution} = X_{solvent}*P_{solvent}$

Where;

$P_{solution$ is the vapor pressure of the solution

$X_{solvent$ is the mole fraction of the solvent

$P_{solvent$ is the vapor pressure of the pure solvent

Thus,

15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl

0.67 kg of water =  [(0.67*1000)/(18)]moles = 37.222 moles of H₂O

Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)

Mole fraction of solvent (water) = (37.222)/(37.222+0.261)

Mole fraction of solvent (water) = 0.993

Note: the vapor pressure of water at 25°C is 0.0313 atm

Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

the vapor pressure of the solution = 0.0311 atm = 23.636 torr

Answer 2

Answer:

0.0310 atm.

Explanation:

The vapor pressure of water at 25°C is 0.0313 atm.

Number of moles:

NaCl

Number of moles = mass/molar mass

Molar mass of NaCl = 23 + 35.5

= 58.5 g/mol.

Moles = 15.27/58.5

= 0.26 mol.

Water:

Molar mass = (1*2) + 16

= 18 g/mol

Number of moles = mass/molar mass

= 670/18

= 37.22 mol.

Using Raoult's law,

Mathematically,

Psol = Xsolv * P°solv

Where

Psol = vapour pressure of the solution

Xsolv = mole fraction of the solvent, water.

P°solv = vapour pressure of the solvent, water

Mole fraction of a substance in a solution is the ratio of its moles by the total number of moles in the solution.

Xa = moles of a/(moles of (a+b)

Xwater = 37.22/0.26+37.22

= 0.993

Psol = 0.993 * 0.0313

= 0.0310 atm