Assuming 100% dissociation, calculate the freezing point and boiling point of 1.22 m sncl₄(aq). constants may be found here.
colligative constants
constants for freezing-point depression and boiling-point elevation calculations at 1 atm:
solvent - formula - value*(°c/m) - normal freezing point (°c) - 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
*when using positive values, assume that ? is the absolute value of the change in temperature. if you would prefer to define ? as "final minus initial" temperature, then ? will be negative and so you must use negative values. either way, the freezing point of the solution should be lower than that of the pure solvent.
= celsius
= celsius

T° freezing solution → -11.3°C

T° boiling solution → 103.1 °C

Explanation:

Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"

This salt dissociates as this:

SnCl₄ (aq)  →   1Sn⁴⁺ (aq)  +   4Cl⁻  (aq)   (so i =5)

The formula for the colligative property of freezing point depression and boiling point elevation are:

ΔT = Kf . m . i

where ΔT = T° freezing pure solvent - T° freezing solution

ΔT = Kb . m . i

where ΔT = T° boiling solution - T° boiling pure solvent

Freezing point depression:

0° - T° freezing solution = 1.86°C/m . 1.22 m . 5

T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C

Boiling point elevation:

T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5

T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

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