Question

An aqueous solution containing 12.50 g of barium chloride in 1000 g of water boils at 373.0834 K. Calculate the degree of dissociation of barium chloride.

Given K_b for H₂O = 0.52 K kg mol⁻¹; molecular mass of BaCl₂ = 208.34 g mol⁻¹.

Answer 1

Given:

\[\ce{W_{BaCl_2}}\] = 12.50 g

\[\ce{W_{H_2O}}\] = 1000 g

T_b = 373.0834 K

K_b = 0.52 K kg/mol

\[\ce{M_{BaCl_2}}\] = 208.34 g mol⁻¹

To calculate, degree of dissociation (α)

ΔT_b = 373.0834 − 373

ΔT_b = 0.0834 K

molatity (m) = `("moles of BaCl"_2)/("Weight of water (kg)")`

= `("W"_("BaCl"_2)//"M"_("Bacl"_2))/(1" Kg")`

= `(12.50//208.34)/1`

= 0.06 m

ΔT_b = = iK_bm, i = `(Δ"T"_"b")/("K"_"b"m")`

= `(0.0834)/(0.52 xx 0.06)`

= 2.67

BaCl₂ − Ba⁺² + 2Cl⁻

l − α   α 2α

i = 1 − α + α 2α

Putting the value of i

2.67 = 1 + 2α

2α = 1.67

α = 0.835

Therefore, the degree of dissociation of Barium chloride is 83.5%.