Question

Enter your answer in the provided box. The formula that governs the depression of freezing point and elevation of boiling point for a solution consisting of a solute dissolved in a solvent is: where:AT = the temperature change between a pure solvent and its solution i = the number of species per mole of solute that are dissolved in the solvent (e.g., i = 1 for a non-ionic solute that does not break apart into ions, i = 2 for an ionic solute such as Br that breaks apart into two ions, K* and Br, and so on) a constant related to the solvent m = molality of the solute (molality, m = moles of solute / kg of solvent) Using the above equation, how many grams of salt (CaCh) would need to be added to 1 L of water in order for the boiling point of the solution to reach 103.5 °C? Assume that the density of water is 1.0 g/mL and that CaCl2 completely dissociates into three ions - ie., a Ca2* ion and two CI ions. The boiling point constant, kb, for water is 0.515 oC/m. Report your answer to the nearest 0.1g.

Answer 1

Since CaCl2 dissociates into 3 ions , i = 3

Mass of 1 L water will be 1 kg as the density is 1.0 g/mL

kb = 0.515^oC/m

Elevation in boiling point = 103.5 - 100 = 3.5^oC

So putting all the values in the formula , we get :

3.5 = 3 x 0.515 x m

m = 1.359 m

Molality = moles of solute / mass of solvent in kg

1.359 = n / 1

n = 1.359

So the mass of CaCl2 added = 1.359 x 110.98

= 150.8 grams