1.

A 0.01m aqueous solution of AlCL_3 freezes at -0.068^@C Calculate the percentage of dissociation.

Answer»

SOLUTION :USE the relation : `Delta T_f=iK_fm`, where I is van.t Hoff factor.
Given that `m=0.01 , Delta T_f=0.068^@C, K_f=1.86K kg mol^-1`
Substituting the values in the above equation, we have
`0.068 k=1 times 86 k kg mol^-1 times 0.01 mol kg^-1`
or `i=0.068/(1.86 times 0.01)3.6559`....(i)
`AlCl_3 to AL^(3+)+3Cl^(-)`
1-x x 3X
`i=(1+3x)/1`
From (i) and (ii), we have
`1+3x=3.6559 or 3x=2.6559 or x=0.8853`
Percentage of dissociation =`0.8853 times 100=88.53`


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