Question

The common laboratory solvent diethyl ether (ether) is often used to purify substances dissolved in it. The vapor pressure of diethyl ether , CH3CH2OCH2CH3, is 463.57 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 7.745 grams of the compound were dissolved in 159.9 grams of diethyl ether, the vapor pressure of the solution was 457.87 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound?

Answer 1

Answer: The molecular weight of this compound is 288.4 g/mol

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute  =$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 7.745 g of compound is present in 159.9 g of diethyl ether

moles of solute = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.745g}{Mg/mol}$

moles of solvent (diethyl ether) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{159.9g}{74.12g/mol}=2.157moles$

Total moles = moles of solute + moles of solvent  = $\frac{7.745g}{Mg/mol}+2.157$

$x_2$ = mole fraction of solute =$\frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}$

$\frac{463.57-457.87}{463.57}=1\times \frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}$

$M=288.4g/mol$

Thus the molecular weight of this compound is 288.4 g/mol