What is the boiling point (in °C) of a 1.56 m aqueous solution of #CaCl_2#?

1 Answer
Jul 15, 2016

Answer:

The boiling point of the solution is #102.39^oC#.

Explanation:

For this question we have to use the boiling point elevation equation:
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Now, let's determine our known and unknown variables:

Known:
molality
Van't Hoff factor
Molal boiling point constant (we're given this value)

Unknown:
Change in boiling point

I should mention that the Van't Hoff factor basically reflects the number of ions produced in solution upon dissociation of an ionic compound. The compound can also be a molecular one, but the Van't Hoff factor for that is just 1 because molecular compounds do not produce ions in solution.

When #CaCl_2# dissociates, you obtain three ions.

#CaCl_2 rarr Ca^(2+) + 2Cl^(-)#

You have one calcium ion and two chloride ions.

Now we can plug in what we know and solve for #DeltaT_b#:

#DeltaT_b = (0.51^oC)/cancelmxx1.56cancelmxx3#

#DeltaT_b = 2.39^oC# This is not your answer. This just tells you how much the boiling point will increase by

Now, we add #100^oC# to the #DeltaT_b#, to determine the new temperature at which the water will boil:

#100^oC + 2.39^oC = 102.39^oC#

***We added #100^oC# to that value because #100^oC# is the boiling point of water. *