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