Answer:
* [tex]\Delta T_B=1.286\°C[/tex]
* [tex]m=2.5m[/tex]
* [tex]n=0.05mol[/tex]
* [tex]M=59.76g/mol[/tex]
Explanation:
Hello,
In this case, considering the boiling point rise problem, we consider its appropriate equation:
[tex]\Delta T_B=imK_b[/tex]
Whereas i is the van't Hoff factor that for this nonvolatile solute is 1, m is the molality, Kb the boiling point constant of water as it is the solvent and ΔT the temperature difference. In such a way, with the given information we obtain:
- ΔT:
[tex]\Delta T_B=101.286\°C-100.000\°C\\\\\Delta T_B=1.286\°C[/tex]
- Molality (mol/kg):
[tex]m=\frac{\Delta T_B}{i*K_b}=\frac{1.286\°C}{1*0.512\°C/m}\\ \\m=2.5m[/tex]
- Moles for 20.02 g (0.02002 kg) of water:
[tex]n=2.5mol/kg*0.02002kg\\\\n=0.05mol[/tex]
- Molar mass:
[tex]M=\frac{mass}{moles}=\frac{3.005g}{0.050mol} \\\\M=59.76g/mol[/tex]
Best regards.
