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

Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.52 m SnCl4(aq). Constants may be found here.

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

vent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride   CCl4 29.8 –22.9 5.03 76.8 camphor   C10H16O 37.8 176

Explanation / Answer

I have solved the first four sub-parts, please post one more question to get the remaining answer

Vont hoff factor for SnCl4 (i) = 5

Depression in Freezing point of water= i * KF * m

=> 5 * 1.86 * 1.52

=> 14.136

Hence freezing point of water = -14.136C

Elevation in boiling point of water= i * Kb * m

=> 5 * 0.512 * 1.52

=> 3.8912

Hence boiling point of water = 103.8912 C

Depression in Freezing point of benzene= i * KF * m

=> 5 * 5.12 * 1.52

=> 38.912

Hence freezing point of benzene = -33.422C

Elevation in boiling point of benzene = i * Kb * m

=> 5 * 2.53 * 1.52

=> 19.228

Hence boiling point of benzene = 99.328 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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