Answer: Molecular weight of this compound is 387.3g/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 : 18.16 g of compound is present in 228.6 g of benzene
moles of solute = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{18.16g}{Mg/mol}[/tex]
moles of solvent (benzene) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{228.6g}{78g/mol}=2.93moles[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{18.16g}{Mg/mol}}{\frac{18.16g}{Mg/mol}+2.93}[/tex]
[tex]\frac{73.03-71.88}{73.03}=1\times \frac{\frac{18.16g}{Mg/mol}}{\frac{18.16g}{Mg/mol}+2.93}[/tex]
[tex]M=387.3g/mol[/tex]
The molecular weight of this compound is 387.3g/mol
