SECTION -C A 0.01 {\m} aqueous solution of {\AlCl}_3 freezes at 0.068^{\circ} {\C}. Calculate the percentage of dissociation. [Given : {\K}_f for Water =1.68 {\K} {\kg} {\mol}^{-1} ]

CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

Question : 28 of 37

Marks: +1, -0

SECTION -C

0.01 {\m}

aqueous solution of

{\AlCl}_3

freezes at

0.068^{∘} {\C}

. Calculate the percentage of dissociation.
[Given :

{\K}_f

for Water

=1.68 {\K} {\kg} {\mol}^{-1}

]

Solution:

Given,

m=0.01 {\m}

\;∆ {\T}_f( {\ s})=-0.068^{∘} {\C}

\; {\K}_f( {\aq})=1.86 {\K} {\kg} {\mol}^{-1}

\;∆ {\T}_f=i {\K}_f m

\;i=\;{∆ {\T}_f}/{ {\K}_f} × m

\;i=\;{0.068}/{1.86} × 0.01 {\m}=3.65

\table ,{\AlCl}_3,→,{\Al}^{3+},+,3 {\Cl}^{-} ; \;\text "initial ",1 {\mol},,0,,0 ; \;\text " At equilibrium " ,1-α,,α,,3 α

Total number of moles at equilibrium

=1-α+α+3 α=1+3 α

l=\;\;{\;\text " Total no. of moles at equilibrium "\;}/{\;\text " Initial no. of moles "\;}

\;=\;{1+3 α}/{1}

3.65\;=1+3 α

α\;=\;{3.65-1}/{3}

Percentage dissociation

=0.88 \%

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