Raoult's Law: Vapor Pressure and Non-Volatile Solutes - Problem #1 (2023)

Raoult's law: The effect of non-volatile solutes on vapor pressureProblems No. 1 - 15

Returning to the discussion of the non-volatile effect on vapor pressure

Problem Nr. 1:What is the vapor pressure of an aqueous solution with a mole fraction of 0.1000? The vapor pressure of water is 25.756 mmHg at 25°C.

Solution:

HSolution= 1.0000 − 0.1000 = 0.9000 <--- solv means solvent

Use Raoult's law:

PAGSolution= (xSolution) ($\text{PAG}{\text{}}_{Solution}^{Ö}$)

x = (0,9000) (25,756)

x = 23.18 mmHg (up to the next four digits)

Problem Nr. 2:The vapor pressure of an aqueous solution is 24.90 mmHg at 25°C. What is the mole fraction of solute in this solution? The vapor pressure of water is 25.756 mm Hg at 25 °C.

Solution:

Use Raoult's law:

PAGSolution= (xSolution) ($\text{PAG}{\text{}}_{Solution}^{Ö}$)

24,90 = (x) (25,756)

x = 0.966765 (this is the mole fraction of the solvent)

Hdissolved substance= 1 − 0,966765 = 0,033235

Hdissolved substance= 0.03324 (up to four nearest digits)

Problem Nr. 3:How many grams of the non-volatile compound B (molar mass = 97.80 g/mol) would need to be added to 250.0 g of water to produce a solution with a vapor pressure of 23.756 torr? The vapor pressure of water at this temperature is 42,362 Torr.

Solution:

We assume that B does not ionize in solution.

1) Determine the mole fraction of solvent that produces a solution vapor pressure of 23.756 torr:

PAGSolution= (xSolution) ($\text{PAG}{\text{}}_{Solution}^{Ö}$)

23,756 Torr = (x) (42,362 Torr)

x = 0,5608

2) Determine the moles of compound B needed to make the mole fraction of the above solvent:

0,5608 = 13,88 / (13,88 + B)

7.7839 + 0.5608B = 13.88

0.5608B = 6.0961

B = 10,87 Mol

3) Determine the mass of B

(10,87 mol) (97,80 g/mol) = 1063 g

Comment: That's a completely ridiculous amount to dissolve in 250.0 g of water, but that's not the point. It's about solving the problem.

Problem Nr. 4:At 29.6 °C, pure water has a vapor pressure of 31.1 torr. A solution is prepared by adding 86.8 grams of "Y", a non-volatile non-electrolyte, to 350 grams of water. The vapor pressure of the resulting solution is 28.6 Torr. Calculate the molar mass of Y

Solution:

1) Use Raoult's law to find the mole fraction of the solvent:

(Video) Solution #3 | Raoult’s Law |Raoult’s Law for volatile solutes | Raoult’s Law for nonvolatile solutes

PAGSolution= (xSolution) ($\text{PAG}{\text{}}_{Solution}^{Ö}$)

28.6 Torr = (χSolution) (31.1 tons)

HSolution= 28,6 Torr / 31,1 Torr

HSolution= 0,91961415

2) Use the mole fraction and moles of water to find the moles of Y:

350 g / 18,015 g/mol = 19,428254 mol

0.91961415 = 19.428254 / (19.428254 + x) <--- x is equal to the amount of Y dissolved

(0,91961415) (19,428254 + x) = 19,428254

17,8665 + 0,91961415x = 19,428254

0,91961415x = 1,561754

x = 1,69827 Mol

3) Calculate the molar mass of Y:

86.8 g / 1.69827 mol = 51.1 g/mol (up to three-digit sig)

Problem Nr. 5:The vapor pressure of pure water is 23.8 mmHg at 25.0 °C. What is the vapor pressure of 2.50 molal C?6H12Ö6

Solution:

1) Convert the molality to a mole fraction. First calculate the total mole:

2,50 mC6H12Ö6= 2,50 mol / 1,00 kg H2Ö

1000 g / 18,015 g/mol = 55,51 mol

55.51 Mol + 2.50 Mol = 58.01 Mol

2) We need the mole fraction of water:

55,51 / 58,01 = 0,9569

3) Use Raoult's law:

PAGSolution= (xSolution) ($\text{PAG}{\text{}}_{Solution}^{Ö}$)

PAGSolution= (0,9569) (23,8 mmHg) = 22,8 mmHg

Problem Nr. 6:How many grams of testosterone, C19H28Ö2, a non-volatile non-electrolyte (MW = 288.4 g/mol), must be added to 207.8 g of benzene to lower the vapor pressure to 71.41 mm Hg? (Benzene = C6H6= 78.12 g/mole. The vapor pressure of benzene is 73.03 mm Hg at 25.0 °C.)

Solution:

Use Raoult's law:

PAGSolution= ($\text{PAG}{\text{}}_{Solution}^{Ö}$) (XSolution)

71,41 = (73,03) (2,66 / (2,66 + x))

71,41 = 194,2598 / (2,66 + x)

71,41 = 194,2598 / (2,66 + x)

194,2598 = 189,9506 + 71,41x

71,41x = 4,3092

x = 0,0603445 Mol

(288.4 g/mol) (0.0603445 mol) = 17.4 g (up to three figs sig)

Problem Nr. 7:At 25.0 °C the vapor pressure of benzene (C6H6) is 0.1252 atm. When 10.00 g of an unknown non-volatile substance is dissolved in 100.0 g of benzene, the vapor pressure of the solution at 25.0 °C is 0.1199 atm. Calculate the mole fraction of the solute in the solution assuming there is no dissociation by the solute.

Solution:

(Video) Finding the Vapor Pressure of a Solution (Ionic-Nonvolatile Solute)

1) Since the solute is non-volatile, the solution vapor contains only benzene. All unknown substances remain in solution. Assuming an ideal mixture, the vapor pressure of the solution is given by Raoult's law:

PAGSolution= ($\text{PAG}{\text{}}_{Solution}^{Ö}$) (XSolution)

2) The mole fraction of benzene in this mixture is:

PAGSolution= ($\text{PAG}{\text{}}_{Petrol}^{Ö}$) (XPetrol)

HPetrol= PAGSolution/$\text{PAG}{\text{}}_{Petrol}^{Ö}$

= 0.1199 atm / 0.1252 atm

= 0,9576677

3) The mole fractions of the components in each mixture add up to one. So for this solution:

HPetrol+hSolution= 1 <--- the 'solu' means solved

For this reason:

HSolution= 1 - xPetrol

= 1 − 0.9576677 = 0.04233 (to four nearest digits)

Problem Nr. 8:What is the vapor pressure at 25.0 °C of a solution consisting of 42.71 g of naphthalene (a non-volatile compound, MW = 128 g/mol) and 40.65 g of ethanol (MW = 46.02 g/mol). ? (The vapor pressure of pure ethanol at 25.0 °C is 96 torr.)

Solution:

1) The vapor pressure of this type of solvent is related to the mole fraction of the solvent and its pure vapor pressure:

solution vapor pressure = (mole fraction of solvent) (vapor pressure of pure solvent)

This is known as Raoult's law.

2) Calculation:

Mol Naphthalin ---> 42,71 g / 128 g/mol = 0,334 mol

Mol Ethanol ---> 40,65 g / 46,02 g/mol = 0,883 mol

Ethanol-Molbruch ---> 0,883 / (0,883 + 0,334) = 0,726

Vapor pressure of solution ---> (96 torr) (0.726) = 70 torr

Problem Nr. 9:A non-volatile organic compound Z was used to prepare a solution. Solution A contains 5.00 g Z dissolved in 100.0 g water and has a vapor pressure of 754.5 mmHg at the normal boiling point of water. Calculate the molar mass of Z

Solution:

1) Use Raoult's law to find the mole fraction of the solvent:

PAGSolution= (xSolution) ($\text{PAG}{\text{}}_{Solution}^{Ö}$)

754.5 Torr = (χSolution) (760,0 Torr)

Note: 760.0 Torr is the vapor pressure of water at its normal boiling point of 100°C.

HSolution= 754,5 Torr / 760,0 Torr

HSolution= 0,99276316

2) Use mole fraction and moles of water to find moles of Z:

100 g / 18,015 g/mol = 5,55093 mol

0.99276316 = 5.55093 / (5.55093 + x) <--- x is equal to the dissolved moles of Z

x = 0,040464 Mol

3) Calculate the molar mass of Z:

5.00 g / 0.040464 mol = 124 g/mol (up to three-digit sig)

Problem Nr. 10:What is the molality of an aqueous solution of urea, CO(NH2)2, if the vapor pressure above the solution is 22.83 mmHg at 25°C? Assume that urea is non-volatile. The vapor pressure of pure water is 23.77 mmHg at 25°C.

Solution:

(Video) raoult's law for volatile and non volatile solutes

PAGSolution= ($\text{PAG}{\text{}}_{Solution}^{Ö}$) (mole fraction of solvent)

22,83 = (23,77) (x)

x = 0.960 (this is the mole fraction of water)

Urea mole fraction = 0.040

Suppose there are a total of 1,000 moles of solvent and solute.

Convert 0.960 moles to grams of water:

(0,960 mol) (18,015 g/mol) = 17,2944 g

Calculate the molality:

0,040 Mol / 0,0172944 kg = 2,31 m

Problem Nr. 11:Calculate the mass of propylene glycol (C3H8Ö2) which must be added to 500. grams of water to reduce the vapor pressure by 4.75 mmHg at 40.0 °C.

Solution:

Seekthe vapor pressure of water at 40.0 °C. It is 55.3 mmHg.

Use Raoult's law to find the mole fraction of the solvent:

50,55 mmHg = (55,3 mmHg) (x)

x = 0,9141

The 50.55 is 55.3 minus 4.75.

The mole fraction of solute is 0.0859

Let's set up a mole fraction calculation:

(x/76.0942) divided by [(x/76.0942) + (500th/18.0152)] = 0.0859

x/76.0942 is the mole of C3H8Ö2, the 500th / 18.0152 is the mole of water and the sum of the two (inside the parentheses) is the total moles in the solution.

solution for x.

Problem Nr. 12:What is the vapor pressure of water over a solution in which 32.5 g of glycerol (C3H8Ö3) dissolve in 125 g of water at 343 K? The vapor pressure of pure water at 343 K is 233.7 torr.

Solution:

The vapor pressure is proportional to the mole fraction in the solution.

Mol Glycerin = 32,5 g / 92,19 g/mol = 0,3525 Mol
Mole water = 125g / 18.0g/mole = 6.944

Total moles = 7.2965 moles
Mole fraction of water = 6.944 moles / 7.2965 moles = 0.9516

(0,9516) (233,7 Torr) = 222,4 Torr

Problem Nr. 13:A solution is prepared by dissolving 396 g sucrose in 624 g water at 30.0°C. What is the vapor pressure of this solution? (The vapor pressure of water is 31.82 mmHg at 30.0 °C.)

Solution:

1) Determine the moles of solute and solvent:

396 g / 342,2948 g/mol = 1,1569 mol
624 g / 18,015 g/mol = 34,6378 mol

2) Determine the mole fraction of the solvent:

34.6378 Mol / (34.6378 Mol + 1.1569 Mol) = 0.96768

3) Determine the vapor pressure:

x = (31,82 mmHg) (0,96768) = 30,8 mmHg (a tres sig feigen)

Problem Nr. 14:Calculate the vapor pressure of a solution made by dissolving 21.80 g glucose (molar mass = 180.155 g/mol) in 460.0 g H2Or at 30.0°C. (The vapor pressure of the pure solvent is 31.82 mmHg at 30.0 °C.)

(Video) NEET/JEE/AIIMS 2019 | Solutions(Vapour Pressure + Raoult Law ) Chemistry (L-8) | by Arvind Arora

Solution:

1) Determine the moles of solute and solvent:

21,80 g / 180,155 g/mol = 0,1210 mol
460,0 g / 18,015 g/mol = 25,5343 mol

2) Determine the mole fraction of the solvent:

25.5343 Mol / (25.5343 Mol + 0.1210 Mol) = 0.9952836

3) Determine the vapor pressure:

x = (31.82 mmHg) (0.9952836) = 31.67 mmHg (up to the next four digits)

Problem Nr. 15:A solution of sodium chloride in water has a vapor pressure of 18.5 Torr at 25°C. What would be the vapor pressure of this solution at 45°C? The vapor pressure of pure water is 23.8 torr at 25°C and 71.9 torr at 45°C.

Solution:

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature multiplied by the mole fraction of the solvent present:

solution at vapor pressure = (mole fraction of solvent) (solvent at vapor pressure)

18,5 Torr = (Molbruch) (23,8 Torr)

solvent mole fraction = 0.77731

A 45°C,

Vapor pressure solution = (0.77731) (71.9 torr) = 55.9 torr (at three SF)

Additional issue:At 300 °C the vapor pressure of Hg is 32.97 torr. What mass of Au would have to be dissolved in 5,000 g Hg to lower its vapor pressure to 25.00 Torr?

Solution #1:

Au dissolves in Hg to form a solution. At 300°C, gold is still solid and has a much lower Au vapor pressure than Hg (which is liquid at 300°C). Therefore, we treat the solute Au as non-volatile.

PAGSolution=$\text{PAG}{\text{}}_{Hg}^{Ö}$· $\text{H}{\text{}}_{Hg}^{}$<--- Mole fraction calculated using vapor pressures

25,00 Torr = (32,97 Torr) ($\text{H}{\text{}}_{Hg}^{}$)

$\text{H}{\text{}}_{Hg}^{}$= 0,758265

Moles of Hg present ---> 5.00 g / 200.59 g/mol = 0.0249265 mol

Let M = moles of dissolved Au.

 molHg $\text{H}{\text{}}_{Hg}^{}$ = ––––––––––– <--- Mole fraction calculated with mol Mol Hg + M
 0,0249265 0,758265 = –––––––––––––– 0,0249265 + M

(0,758265) (0,0249265 + M) = 0,0249265

0,0189009 + 0,758265M = 0,0249265

0,758265 M = 0,0060256

M = 0,00794656 mol Au

(196.96657 g/mol) (0.00794656 mol) = 1.565 g (up to 4 digits Sig)

Solution #2:

This method corresponds to the two methods for calculating mole fraction.

$\text{H}{\text{}}_{Hg, with vp}^{}$=$\text{H}{\text{}}_{Hg, with Mol}^{}$

 PAGSolution molHg ––––––– = ––––––––––– $\text{PAG}{\text{}}_{Hg}^{Ö}$ Mol Hg + M

sea ​​m = mass of Au

25.00
 5.00 ––––– 200.6
–––––=–––––––––––––––
32.97
 5.00 Metro –––––  + ––––– 200.6 197.0

Note: Molar masses have been rounded and all units have been removed.

0,0249265
0,758265=–––––––––––––––––––
 Metro 0,0249265  + ––––– 197.0
 0,758265 m 0,01890089  + ––––––––––  = 0,0249265 197.0
 0,758265 m ––––––––––  = 0,00602561 197.0

0,758265 m = 1,18704517

m = 1.565 g (up to four figs)

Returning to the discussion of the non-volatile effect on vapor pressure

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