Question

The specific conductance of 0.01 m solution of kcl is 0.0014 1 cm1 at 25c. Its equivalent conductance (cm2 1eq1) is

Answer 1

Specific conductance (k)=0.0014 ohm^-1 cm^-1
Molarity of kcl = 0.01M
Now...we know that
Equivalent conductance= k×1000/normality
Normality of 0.01 M kcl is 0.01N.........bcz n=1.........for kcl
Hence,. Equivalent conductance = 0.0014×1000/0.001
= 140 cm^2ohm^-1eq^-1

Answer 2

Answer:

140 $cm^{2}ohm^{-1}eq^{-1}$

Explanation:

Specific conductance (k)=0.0014 $ohm^{-1}cm^{-1}$

Molarity of KCl = 0.01M

Equivalent conductance= $\frac{k(1000)}{normality}$

Normality of 0.01 M KCl is 0.01N since n=1 for KCl

Hence,

    Equivalent conductance = $\frac{(0.0014)1000}{0.001}$

= 140 $cm^{2}ohm^{-1}eq^{-1}$

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