Question

Determine the boiling point of a solution made by dissolving 0.60 mol K2SO4 in 1.0 kg water. Water has a boiling point elevation constant of 0.51°C•kg/mol. What is the boiling point of this solution?

Answer 1

to determine the boiling point (bp), you need this formula---> bp solution= bp solvent + Δbp solution

bp solvent is known since we are dealing with water; it is 100C

all we have to do is to solve for Δbp solution and add it to the 100C. 

Δbp solution= Kb x i x m, where Kb is the constant, i is the number of particles and m is molality. 

Kb= 0.51 
i= 3, K2SO4 ----> 2K+ + SO4-2 (the solute breaks into three ions)
m= ? = moles/kg of solvent

molality (m)= 0.60 mol/ 1.0 Kg= 0.60 m

Δbp solution= 0.51 x 3 x 0.60= 0.918 C

bp solution= 100C + 0.918 C= 100.918 C

Answer 2

Answer:

1) -65.9

2) 100.31

Explanation: edg 2020