Question

The conductivity of 0.02M AgNO3 at 25°C is 2.428×10^-3.What is its molar conductivity? ​

Answer 1

Answer:

Molar Conductivity = 121.4 Ω⁻¹ cm² mol⁻¹

Explanation:

We Have :-

C = 0.02M AgNO₃

T = 25 °C

Conductivity ( k ) = 2.428 * 10⁻³ Ω⁻¹cm⁻¹

To Find :-

Molar Conductivity

Fomula Used :-

Molar Conductivity = 1000k / C

Solution :-

$Molar\ Conductivity=\frac{1000\ k}{C}\\\\= \frac{1000*2.428*10^{-3} }{0.02} \\\\= 121.4$

Molar Conductivity = 121.4 Ω⁻¹ cm² mol⁻¹

Answer 2

Answer:

Molar Conductivity = 121.4 0-¹ cm² mol

Explanation:

We Have :

C=0.02M AgNO3

T= 25 °C

Conductivity (k)=2.428*10-³

To Find :

Molar Conductivity

Fomula Used :

Molar Conductivity=1000k/C

Solution :

1000 k C Molar Conductivity

1000 * 2.428 * 10-³

0.02

=121.4

Molar Conductivity = 121.40-¹

cm² mol-1

Explanation:

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