Question

Find the degree of dissociation when 0.02 M aqueous Al2(SO4)3 solution having elevation of boiling point of 0.04k the molal elevation constant for solvent is 1.9k kg mol-¹

Answer 1

degree of dissociation is 1.31% (approximately).

first we have to find Van't Hoff factor , i

using formula, ∆T_b = i × K_e × m

where ∆T_b is elevation in boiling point, K_e is molal elevation constant and m is molality.

here, ∆T_b = 0.04K, K_e = 1.9 K kg/mol

and molality, m = 0.02m

now, 0.04 = i × 1.9 × 0.02

⇒i = 2/1.9 = 1.0526

dissociation of Al2(SO4)3 ;

Al2(SO4)3 ⇔2Al³+ + 3SO4²-

at t = 0, α 0. 0

at eq 1 - α 2α 3α

so, i = 1 - α + 2α + 3α = 1 + 4α

⇒1.0526 = 1 + 4α

⇒α = 0.0526/4 ≈ 0.0131

hence, % degree of dissociation = 100α= 1.31 %

also read similar questions : An aqueous solution of glucose boils at 100.01 degree celsius. The molal boiling point elevation constant for water is 0...

https://brainly.in/question/5084354

mole fraction of solvent in aqueous solution of NaOH having molality of 3 is

https://brainly.in/question/3601574