If am for BaCl2, NaOH and NaCl are a, b and
c 0-1 cm2 mol-1 respectively, the
equivalent conductivity of Ba(OH)2 at
infinite dilution is
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Explanation:
∧
m
o
(NaOH)=λ
Na
+
o
+λ
OH
−
o
∧
m
o
(NaCl)=λ
Na
+
o
+λ
a
−
o
∧
m
o
(BaCl
2
)=λ
Ba
+2
o
+2λ
a
−
o
⇒2∧
m
o
(NaOH)−2∧
m
o
(NaCl+∧
m
o
(BaCl
2
)
2λ
Na
+
o
+2λ
OH
−
−(2λ
Na
+
o
+2λ
a
−
o
)+λ
Ba
+2
o
+2λ
a
−
o
λ
m
o
Ba(OH)
2
=λ
Ba
+2
o
+2λ
OH
−
So ∧
m
o
=(2×248−2×126.5+280)×10
−4
∧
m
o
Ba(OH)
2
=523×10
−4
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