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An aqueous solution containing 12.48g of...

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 `K_(b)` for `H_(2)O =0.52 Km^(-1)`, Molar mass of `BaCl_(2) =208.34 "g mol"^(-1)`]

Text Solution

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Elevation of boiling point is given by `DeltaT_(b)= iK_(bm`
Where, i is the van.t Hoff factor, `K_(b)` is freezing point constant and m is the molality of solution
Given:
`K_(b)=0.52" K kg mol"^(-1)`
`DeltaT_(b)=373.233K-373.15K=0.083K`
Molarity, `m=([(12.48)/(207)])/(1)`
`=0.06"mol kg"^(-1)`
Putting the values in equation :
`0.083K =i xx 0.52 "K kg mol"^(-1) xx 0.06 "mol kg"^(-1)`
`i =(0.083)/(0.0312)`
`=2.67`
Now, `underset(1-x)(BaCl_(2)) to underset(x)(Ba^(+)) + underset(2x)(2Cl^(-))`
(where x is the degree of dissociation of `BaCl_(2)`)
The total number of moles after dissociation
`=1-x+x+2x=1+2x`
`i =("Normal colligative property")/("Abnormal colligative property")`
`2.67=(1+2x)/(1)`
`1+2x =(2.67)`
`x=(1.67)/(2)=0.835`
Degree of dissociation is 0.835 of 83.5%.
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