0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 m m. The molar conductivity of the solution will be ______ mS . (Nearest integer) [2025]

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Solution

Q.
0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 m $Ω$ m. The molar conductivity of the solution will be ______ $\times 10^{2}$ mS ${dm}^{2}$ ${mol}^{- 1}$ . (Nearest integer) [2025]

Ans.
(23)

$Molarity of NaOH (C) = \frac{(% w / v) \times 10}{M_{NaOH}}$

$= \frac{0.2 \times 10}{40} = 0.05 M$

$ρ = 870 mΩ m = 870 \times 10^{- 3} Ω \times 100 cm = 87 Ω cm$

$κ = \frac{1}{ρ} = \frac{1}{87} {S cm}^{- 1}$

$Λ_{m} = \frac{1000 κ}{C} {S cm}^{2} {mol}^{- 1}$

$= \frac{1000}{87 \times 0.05} {S cm}^{2} {mol}^{- 1} = 229.885 {S cm}^{2} {mol}^{- 1}$

$= 229.885 \times 1000 {mS (10}^{- 1} {dm)}^{2} {mol}^{- 1}$

$= 22.99 \times 10^{2} {mS dm}^{2} {mol}^{- 1}$

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