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

1 Answer
Jul 15, 2016

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. *