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Benzene boils at 80.10 °C and has a molal boiling constant, k b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 °C. What is the molality of the solute?

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sebassandin

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.