Discussion Forum



1)

The resistance of $\frac {1}{10}$ M solution is 2.5 x 103 ohm . What is the molar conductivity of solution ? (cell constant =1.25 cm-1)


A) 3.5 $ohm^{-1}cm^{2}mol_{-1}$

B) 5.0 $ohm^{-1}cm^{2}mol_{-1}$

C) 2.5 $ohm^{-1}cm^{2}mol_{-1}$

D) 2.0 $ohm^{-1}cm^{2}mol_{-1}$

Answer:

Option B

Explanation:

 Given, Molarity = $\frac {1}{10}$ M

               R= 2.5 x 103 $\Omega$

 Cell constant   $(\frac{l}{a})=1.25 cm^{-1}$

 Now, k( conductivity)=   $\frac{1}{k}\times\frac{l}{a}$

 $\therefore$    $k=\frac{1}{2.5\times 10^{3}}\times 1.25=0.5\times 10^{-3}\Omega^{-1}cm^{-1}$

 $ \Rightarrow$  Molar conductivity $(\wedge_{m})=\frac{k\times1000}{M}$

 $\therefore$       $\wedge_{m}=\frac{0.5\times 10^{-3}\times1000}{1/10}$

     $\wedge_{m}=50\Omega^{-1}cm^{2}mol^{-1}$