Question

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.

Answer 1

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 = $p_s$

Mass of glycerin = 92.1 g

Moles of glycerin = $n_1=\frac{92.1 g}{92 g/mol}=1.001 mol$

Mass of ethanol = 184.4 g

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

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

$\frac{p-p_s}{p}=\chi_{solute}$

$\frac{p-p_s}{p}=\chi_{2}=\frac{n_2}{n_1+n_2}$

$\frac{0.178 atm-p_s}{0.178 atm}=\frac{1.001 mol}{1.001 mol+4.009 mol}$

$p_s=0.142 atm$

The vapor pressure of the solution is 0.142 atm.

Weight percentage of glycerin is :

$=\frac{92.1 g}{ 92.1 g+184.4 g}\times 100=33.31\%$

The weight percentage of Glycerin is 33.31%.