Question

6. You dissolve 0.89 g of an organic alcohol into 93.5 g of deionized water to make an aqueous solution. The solution boils at a temperature of 100.32 °C. Use this data to ealeulate the molar mass (g/mol) of the acid. Also state two actions you could take to raise the boiling point even higher and explain each decision briefly. Hint: work backwards from AT5 to find mol solute... Continue your work onto the top of the next page. For water, Kb = 1.86 °C/m 112 pts.

Please show detail of all steps

Answer 1

A Tg = Kb Xm where A Tb = Elevation in boiling point - Is- To where Te = Boiling point of solution = 100. 32" - To= Boiling point of solvent = 100.C - (water) io sto = 100.32°C – 100.00°C = 0.32°C 0 Also, m= Molality of solution = No of moles of solute - kg of solvent = Mass of solute x 1000 Molar Mass of solute grams of solvent 100 mo 0.89g X 1000 - Molar Mass of solute X 93.5g Also, Kb = 1.86°/m I veing ④, Q. ③ in Egn ② 0.3.206 = 1.86°C/m X 0.898 x 1000 - Molar mass X 93.59 Et Molar Mass of solute = 55.3 g/mol do As Elevation in Boiling point is directly proportional to molality and molality i directly proportional I to mass of solute and inversely proportional to mass of solvent. Increasing the mass of solute and/or decreasing the mass of solvent will raise boiling point even higher.