Answer:
[tex]m=0.395mol/kg[/tex]
Explanation:
Hello,
This is a problem about boiling point elevation which is modeled via:
[tex]\Delta T=i*m*Kb[/tex]
Whereas for this solvent (nonpolar, nonionizing), the van't Hoff factor is one. In such a way, the molality of the solute is simply computed as shown below:
[tex]m=\frac{\Delta T}{Kb}=\frac{(81.10-80.10)\°C}{2.53\°C/m} \\\\m=0.395mol/kg[/tex]
In this manner, we can also compute the molar mass of the solute by noticing 20.0 g (0.020 kg) of benzene were used:
[tex]n=0.395mol/kg*0.020kg=7.9x10^{-3} mol[/tex]
And considering the 2.15 g of the solute:
[tex]Molar\ mass=\frac{2.15g}{7.9x10^{-3}mol}\\ \\Molar\ mass=271.975g/mol[/tex]
Best regards.
