The molar conductance at infinite dilution of AgNO3 , AgCl and NaCl are 116.5, 121.6 and 110.3
respectively. Calculate the molar conductances of NaNO3.
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0
Answer:
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Explanation:
Answered by
1
Answer:
Λ
m
∞
(AgNO
3
)=116.5 S cm
2
mol
−1
Λ
m
∞
(AgCl)=121.6 S cm
2
mol
−1
Λ
m
∞
(NaCl)=110.3 S cm
2
mol
−1
Λ
m
∞
(AgNO
3
)=Λ
∞
Ag
+
+Λ
∞
NO
3
−
=116.5.....(i)
Λ
m
∞
(AgCl)=Λ
∞
Ag
+
+Λ
∞
Cl
−
=121.6.....(ii)
Λ
m
∞
(NaCl)=Λ
∞
Na
+
+Λ
∞
Cl
−
=110.3.....(iii)
(i)+(iii)−(ii), we get-
Λ
m
∞
(NaNO
3
)=Λ
m
∞
AgNO
3
+Λ
m
∞
NaCl− Λ
m
∞
AgCl
=Λ
∞
Na
+
+Λ
∞
NO
3
−
=116.5+110.3−121.6
=105.2
The molar conductance of NaNO
3
is 105.2 S cm
2
mol
−1
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