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An ideal solution containing 92.1 g of glycerin, C3H5(OH)3, and 184.4 g of ethanol, C2H5OH, is at 40°C. The vapor pressure of pure ethanol is 0.178 atm at 40°C. Glycerin is essentially nonvolatile at this temperature. Compute the vapor pressure and weight percentage of Glycerin.

Respuesta :

Answer:

The vapor pressure of the solution is 0.142 atm.

The weight percentage of Glycerin is 33.31%.

Explanation:

Vapor pressure of pure ethanol = p = 0.178 atm

Vapor pressure of the glycerin solution = [tex]p_s[/tex]

Mass of glycerin = 92.1 g

Moles of glycerin = [tex]n_1=\frac{92.1 g}{92 g/mol}=1.001 mol[/tex]

Mass of ethanol = 184.4 g

Moles of ethanol = [tex]n-2=\frac{184.4 g}{46 g/mol}=4.009 mol[/tex]

Relative lowering in the vapor pressure of the solution with non volatile solute  is equal to the mole fraction of solute.

[tex]\frac{p-p_s}{p}=\chi_{solute}[/tex]

[tex]\frac{p-p_s}{p}=\chi_{2}=\frac{n_2}{n_1+n_2}[/tex]

[tex]\frac{0.178 atm-p_s}{0.178 atm}=\frac{1.001 mol}{1.001 mol+4.009 mol}[/tex]

[tex]p_s=0.142 atm[/tex]

The vapor pressure of the solution is 0.142 atm.

Weight percentage of glycerin is :

[tex]=\frac{92.1 g}{ 92.1 g+184.4 g}\times 100=33.31\%[/tex]

The weight percentage of Glycerin is 33.31%.

Lanuel

1. The vapor pressure of glycerin is equal to 0.1424 atm.

2. The weight percentage of glycerin is equal to 33.31%.

Given the following data:

  • Mass of glycerin = 92.1 grams
  • Mass of ethanol = 184.4 grams
  • Temperature = 40°C.
  • Vapor pressure of pure ethanol = 0.178 atm.

Scientific data:

  • The molar mass of ethanol = 46.07 g/mol
  • The atomic weight of glycerin = 92.09 g/mol

To calculate the vapor pressure and weight percentage of glycerin:

First of all, we would determine number of moles of each substance.

For glycerin:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{92.1}{92.09}[/tex]

Number of moles = 1.0 moles

For ethanol:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{184.4}{46.07}[/tex]

Number of moles = 4.0 moles

Note: Relative lowering of pressure of an ideal solution with non-volatile solute is equal to the mole fraction of its solute.

Mathematically, this is given by the formula:

[tex]\frac{P^o -P_s}{P^o} = \frac{n_e}{n_e +n_g}[/tex]

Where:

  • [tex]P_s[/tex] is the vapor pressure of solute.
  • [tex]P^o[/tex] is the vapor pressure of solvent.
  • [tex]n_g[/tex] is the number of moles of glycerin.
  • [tex]n_e[/tex] is the number of moles of ethanol.

Substituting the given parameters into the formula, we have;

[tex]\frac{0.178 -P_s}{0.178} = \frac{1}{1\; +\;4}\\\\\frac{0.178 -P_s}{0.178} = \frac{1}{5}\\\\5(0.178 -P_s)=0.178\\\\0.89-5P_s=0.178\\\\5P_s=0.89-0.178\\\\5P_s=0.712\\\\P_s=\frac{0.712}{5} \\\\P_s = 0.1424\;atm[/tex]

Vapor pressure of glycerin = 0.1424 atm.

Lastly, we would compute the weight percentage of glycerin:

[tex]Total \;mass = 184.4 +92.1 \\\\Total \;mass = 276.5\;grams[/tex]

Mathematically, weight percentage is given by the formula:

[tex]Weight\;percent = \frac{Mass}{Total\;mass} \times 100\\\\Weight\;percent = \frac{92.1}{276.5} \times 100\\\\Weight\;percent = \frac{9210}{276.5}[/tex]

Weight percentage = 33.31%

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