Answer: 386.0 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,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : 21.47 g of compound X is present in 233.8 g of diethyl ether
moles of solute (X) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{21.47g}{Mg/mol}[/tex]
moles of solvent (diethyl ether) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{233.8g}{74g/mol}=3.160moles[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]
[tex]\frac{463.57-455.55}{463.57}=1\times \frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]
[tex]0.017301=1\times \frac{\frac{21.47g}{Mg/mol}}{\frac{21.47g}{Mg/mol}+3.160}[/tex]
[tex]M=386.0g/mol[/tex]
The molecular weight of this compound is 386.0 g/mol
