LKOH = lK+ + lOH- [L=molar conductivity, lK+ and lOH- molar ionic conductivity of K+ and OH- respectively]
= 73.5+198.6 Sm2mol-1
= 272.1 Sm2mol-1
L = (1000×K)/C K = conductivity
Conductivity of 0.1 M KOH is
K = 272.1× 0.1/1000 Sm-1
= 0.027 Sm-1 Answer
LNaOH = lNa+ + lOH-
= 50.1+198.6 Sm2mol-1
= 248.7 Sm2mol-1
Conductivity of 0.1 M NaOH is
K = 248.7× 0.1/1000 Sm-1
= 0.024 Sm-1 Answer
LHNO3 = lH+ + lNO3-
= 349.8+71.4 Sm2mol-1
= 421.2 Sm2mol-1
Conductivity of 0.1 M HNO3 is
K = 421.2× 0.1/1000 Sm-1
= 0.042 Sm-1 Answer
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