Question

The electrical resistance of a column of 0.05 M
KOH solution of diameter 1 cm and length 45.5 cm
is 4.55 10 ohm. Calculate its molar conductivity.​

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 × 10³ ohm. Calculate its molar conductivity

$\huge\underline{\underline{\bf \green{Solution-}}}$

• C = 0.05 M
• Diameter ( d ) = 1 cm Radius (r) = 0.5cm
• Lenght ( l ) = 45.5 cm
• Resistance ( R ) = 4.55 × 10³ Ω

$\large\underline{\underline{\sf To\:Find:}}$

• Molar conductivity ${\sf ( \triangle _m)}$

✪$\large{\boxed{\bf \blue{R =\rho \dfrac{l}{A} }}}$

A = Area = πr²

⠀⠀⠀⠀⠀⠀= 3.14 × (0.5)²

⠀⠀⠀⠀⠀⠀= 0.785 cm²

$\implies{\sf 4.55×10^3=\rho \dfrac{45.5}{0.785} }$

$\implies{\sf \rho = \dfrac{4.55×10^3×0.785}{45.5} }$

$\implies{\sf \rho = \dfrac{3.57×10^3}{45.5}}$

$\implies{\bf \rho = 78.5\:Ω.cm }$

✪$\large{\boxed{\bf \blue{Conductivity (K) = \dfrac{1}{\rho}}}}$

$\implies{\sf \dfrac{1}{78.5} }$

$\implies{\bf K = 0.0127\:Scm^{-1} }$

✪$\large{\boxed{\bf \blue{\triangle _m = \dfrac{K × 1000}{C}} }}$

$\implies{\sf \dfrac{0.0127×1000}{0.05} }$

$\implies{\sf \dfrac{12.7}{0.05}}$

$\implies{\bf \red{Molar\: Conductivity (\triangle _m)=254\:Scm^2mol^{-1}}}$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Molar Conductivity ${\bf \red{Molar\: Conductivity (\triangle _m)=254\:Scm^2mol^{-1}}}$

Answer 2

Explanation:

$\huge\bold\red{ANSWER}$

The electrical resistance of a column of 0.05 M KOH solution of di The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55×103ohm. Calculate its molar conductivity.

Answer : 254.8 S cm2mol-1.

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