Answer to Question #333756 in General Chemistry for Jordy Enamorado

Question

If I add 65 grams of calcium chloride to 500 grams of water, what will the boiling

points be of the resulting solution? Kb = 0.512 °C.

Answer 1

CaCl₂ —> Ca²⁺ + 2Cl^-

ΔT_b= i^. K_b⋅b, where

i - the van't Hoff factor;

K_b - the ebullioscopic constant of the solvent;

b - the molality of the solution. 

ΔT_b - the poiling point elevation - defined as T_b − T_b°

The ebullioscopic constant of water is equal to 0.512^∘C kg mol^-1

To determine the molality of the solution, use calcium chloride's molar mass to help you find the number of moles you have in that 65-g sample.

n(CaCl₂) = m / Mr = 65 g / 110.98 g/mol = 0.59 moles

This means that the molality of the solution will be - do not forget to convert the mass of the solvent from grams to kilograms

b = n_solute / m_solvent

b = 0.59 moles / 0.500 kg = 1.18 molal

i (CaCl₂) = 3

ΔT_b= 3 •0.512^∘C kg mol^-1• 1.18 molal= 1.81°C

ΔT_b = T_b - T_b°

T_b = 100 °C + 1.81°C = 101.81°C