Q.
Van't Hoff factor of 80% dissociated BaCl2 solution is
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a
2.2
b
2.6
c
1.8
d
2.4
answer is B.
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Detailed Solution
Examples…..M2SO4(Na2SO4, K2SO4 etc), M2CO3(Na2CO3, K2CO3 etc) ;They all are classified as ‘A2B’ type electrolytes…..A2B ⇌2A++B2- ; n=3 Van't Hoff factor, i = 1-α+nα ; when n= 3 Van't Hoff factor, i = 1+2α ; n=3….for AB2(bi-univalent electrolyte)…...AB2 ⇌A2++2B- ; n=3Examples…..MX2(BaCl2, CaCl2 etc), M(NO3)2 (Ba(NO3)2, Ca(NO3)2 etc)α= Degree of ionization (or) dissociation ;Ex ; For 5% ionized electrolyte α=5100= 0.05 ;0.1, 0.15, 0.2, 0.25, 0.3…..for 10%, 15%, 20%, 25%, 30% ionized electrolyteα for x% electrolyte = x100; For AB2 or A2B type : i= 1+2α ;i= a) 1.1….5% ionized b) i= 1.2…..10% ionized c) i=1.3….15% ionized d) i= 1.4…..20% ionized…..Hint: For every 5% increase in ionization i.e for every 0.05 units increase in 'α' value ‘i’ increases by 0.1 units ;
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