Question

how many grams of salt ( CaCl2 ) would need to be added to 1 L of water in order for the boiling point of the solution to reach 103 °C ? Assume that the density of water is 1.0 g/mL and that CaCl2 completely dissociates into three ions — i.e., a Ca2+ ion and two Cl− ions. The boiling point constant, kb , for water is 0.515°C/m .

Report your answer to the nearest 0.1 g.

Answer 1

The mass of the calcium chloride solute that is dissolved is  215.3 g.

What is the mass of the solute required?

We know that boiling point is one of the colligative properties. It depends on the amount of the substances that is present in the system. Now we have to be able to obtain the solute required to increase the boiling point to  103 °C.

Boiling point elevation = Boiling point of solution - Boiling point of solvent

=  103 °C - 100 °C = 3 °C

ΔT = K m i

ΔT =  Boiling point elevation

m = molality of the solution

i = Van't Hoff factor

3 °C=  0.515°C/m * m * 3

m = 3 °C/0.515°C/m * 3

m = 1.94 m

Molality = mass/molar mass * 1/mass of the solution in kilogram

Since the density of water is 1.0 g/mL and the volume is 1L or 1000 mL then the mass of the water is 1000g of 1Kg

1.94 = mass/111  * 1/1

1.94 = mass/111

mass = 1.94 * 111

mass = 215.3 g

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