Assuming 100% dissociation, calculate the freezing point and boiling point of 1.

Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.85 m Na2SO4(aq). Constants may be found here. Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf value* (

Explanation / Answer

i = 3 .......For solvent as water ......Delta Tf = 3 x 1.86 x 1.85 = 10.323 C ......=> Freezing Point = -10.323 degree C ........Delta Tb = 3 x 0.512 x 1.85 = 2.842 C ......=> Boiling Point = 102.842 C ............For Solvent as Benzene .......Delta Tf = 3 x 5.12 x 1.85 = 28.416 C ......=> Freezing Point = 5.49 - 28.416 = -22.926 degree C .......Delta Tb = 3 x 2.53 x 1.85 = 14.04 C .....=> Boiling Point = 80.1 + 14.04 = 94.14 degree C .............For Solvent as cyclohexane .....Delta Tf = 3 x 20.8 x 1.85 = 115.44 C ......=> Freezing Point = 6.59 - 115.44 = -108.85 C ........Delta Tb = 3 x 2.92 x 1.85 = 16.21 .......=> Boiling Point = 96.91 C ............For Ethanol .....Delta Tf = 3 x 1.99 x 1.85 = 11.04 C .............=> Tf = -128.34 C ....................Delta Tb = 3 x 1.22 x 1.85 = 6.77 ..............=> Tb = 85.17 C ...................For CCl4 ...................Delta Tf = 3 x 29.8 x 1.85 = 165.39 C .................=> Tf = -188.29 C ............Delta Tb = 3 x 5.03 x 1.85 = 27.92 ..........=> Tb = 104.72 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.