Answer:
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![\Delta T_B=1.286\°C](https://img.qammunity.org/2021/formulas/chemistry/college/jd1lgwstourrjdqingej7k7z9wul0feugx.png)
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![m=2.5m](https://img.qammunity.org/2021/formulas/chemistry/college/zjb0y6549k6hb3czcf3q4chwn62w93jcu4.png)
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![n=0.05mol](https://img.qammunity.org/2021/formulas/chemistry/college/kcudlueujy56ui7cljsyq1i1q4hue9cv80.png)
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![M=59.76g/mol](https://img.qammunity.org/2021/formulas/chemistry/college/zeog1vtmct62y5ffdslr5cuxmwslbazanp.png)
Step-by-step explanation:
Hello,
In this case, considering the boiling point rise problem, we consider its appropriate equation:
![\Delta T_B=imK_b](https://img.qammunity.org/2021/formulas/chemistry/college/j9b6d10293q9yypbblpj8xd3hqjgiso0zb.png)
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:
![\Delta T_B=101.286\°C-100.000\°C\\\\\Delta T_B=1.286\°C](https://img.qammunity.org/2021/formulas/chemistry/college/zfz8gv7cgyhv0joxw9zbl5dhpa8dqj0q7s.png)
- Molality (mol/kg):
![m=(\Delta T_B)/(i*K_b)=(1.286\°C)/(1*0.512\°C/m)\\ \\m=2.5m](https://img.qammunity.org/2021/formulas/chemistry/college/p2md754c43ax2zudnydwfzggcp6kdgfjyw.png)
- Moles for 20.02 g (0.02002 kg) of water:
![n=2.5mol/kg*0.02002kg\\\\n=0.05mol](https://img.qammunity.org/2021/formulas/chemistry/college/png1j8piqiev453bz8osff2mxvrbu8htwr.png)
- Molar mass:
![M=(mass)/(moles)=(3.005g)/(0.050mol) \\\\M=59.76g/mol](https://img.qammunity.org/2021/formulas/chemistry/college/czl54e8t0onydiyx9t7e73thi2uociht0f.png)
Best regards.