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(B) Given: Concentration of $NaOH = 0.2 \% (w/v)$.This means $0.2 \ g$ of $NaOH$ is present in $100 \ mL$ of solution.Molarity $(M) = \frac{	ext{mole

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$23$

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(B) Given: Concentration of $NaOH = 0.2 \% (w/v)$.This means $0.2 \ g$ of $NaOH$ is present in $100 \ mL$ of solution.Molarity $(M) = \frac{	ext{moles of solute}}{	ext{Volume of solution in } L} = \frac{0.2 / 40}{100 / 1000} = \frac{0.005}{0.1} = 0.05 \ M$.Given resistivity $(ho) = 870.0 \ m\Omega \ m = 870.0 	imes 10^{-3} \ \Omega \ m = 0.87 \ \Omega \ m$.Since $1 \ m = 10 \ dm$,$ho = 0.87 \ \Omega \ m = 0.87 \ \Omega \ (10 \ dm) = 8.7 \ \Omega \ dm$.Conductivity $(\kappa) = \frac{1}{ho} = \frac{1}{8.7} \ S \ dm^{-1}$.Molar conductivity $(\Lambda_m) = \frac{\kappa}{M} = \frac{1 / 8.7}{0.05} = \frac{1}{8.7 	imes 0.05} = \frac{1}{0.435} \approx 2.2988 \ S \ dm^2 \ mol^{-1}$.Converting to $mS \ dm^2 \ mol^{-1}$: $2.2988 \ S \ dm^2 \ mol^{-1} = 2298.8 \ mS \ dm^2 \ mol^{-1} = 22.988 	imes 10^2 \ mS \ dm^2 \ mol^{-1}$.Rounding to the nearest integer,we get $23 	imes 10^2 \ mS \ dm^2 \ mol^{-1}$.

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

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