An aqueous solution containing 12.48 g of barium chloride

Asked by admin @ 23/08/2021 in Chemistry viewed by 326 People

An aqueous solution containing 12.48g of barium chloride in 1.0 kg of water boils at 373.0832 K. calculate the degree of dissociation of barium chloride. (given, Kb for H2O=0.52 Kmolâ -1, molar mass of BaCl2=208.34 g mol-1)

Answered by admin @ 23/08/2021

Answer: Degree of dissociation is 0.0885

Explanation:-

Weight of solvent = 1.0 kg

Molar mass of $BaCl_2$ = 208.34 g/mol

Mass of $BaCl_2$ added = 12.48 g

$\Delta T_b=i\times K_b\times \frac{\text{mass of barium chloride}}{\text{molar mass of barium chloride}\times \text{weight of solvent in kg}}$

$\Delta T_b=T_f-T^{0}_f$ = elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

$(373.0832-373)K=i\times 0.52\times \frac{12.4}{208.34\times 1.0}$

$0.0832$ =i\times 0.03[/tex]

$i=2.77$

Vant hoff's factor(i) is related to degree of dissociation $(\alpha)$ as:

$\alpha =\frac{i-1}{n-1}=\frac{2.77-1}{3-1}$

$\alpha=0.885$

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