What will be the percentage disassociation of Ca(NO3)2 if , i= 2.5
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The percentage is 1788.8mm
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i= 1 + a(n-1)
2.5=1+a(3-1)
a=1.5/2
%a=1.5/2 x100
=75%
here n = no. of particles formed after dissociation
Ca(NO3)2---> Ca+2 + 2NO3-
therefore n=3 as 1 Ca and 2 No3 forms
2.5=1+a(3-1)
a=1.5/2
%a=1.5/2 x100
=75%
here n = no. of particles formed after dissociation
Ca(NO3)2---> Ca+2 + 2NO3-
therefore n=3 as 1 Ca and 2 No3 forms
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