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

<p>Determine the conductivity from the resistivity. </p> <p><p>The resistivity is given as </p> <p>$\rho = 870.0\; \text{m}\Omega\cdot\text{m}.$ </p></p> <p><p>Convert this into ohm·meters: </p> <p>$$870.0\; \text{m}\Omega\cdot\text{m} = 870.0 \times 10^{-3}\; \Omega\cdot\text{m} = 0.87\; \Omega\cdot\text{m}.$$</p></p> <p><p>Conductivity is the reciprocal of resistivity: </p> <p>$\kappa = \frac{1}{\rho} = \frac{1}{0.87} \approx 1.15\; \text{S/m}.$</p></p> <p>Find the concentration of NaOH in mol/m³. </p> <p><p>A $0.2\%\;(\mathrm{w}/\mathrm{v})$ solution means there are 0.2 grams of NaOH per 100 mL of solution. </p></p> <p><p>In 1 liter (1000 mL) there are: </p> <p>$\frac{0.2\; \text{g}}{100\; \text{mL}} \times 1000\; \text{mL} = 2\; \text{g/L}.$</p></p> <p><p>The molar mass of NaOH is approximately $40\; \text{g/mol}$. Thus, the molarity is: </p> <p>$$\text{Molarity} = \frac{2\; \text{g/L}}{40\; \text{g/mol}} = 0.05\; \text{mol/L}.$$</p></p> <p><p>Since $1\; \text{L} = 0.001\; \text{m}^3$, converting to SI units (mol/m³): </p> <p>$$c = 0.05\; \text{mol/L} \times \frac{1}{0.001\; \text{m}^3/\text{L}} = 50\; \text{mol/m}^3.$$</p></p> <p>Calculate the molar conductivity in SI units. </p> <p><p>Molar conductivity $\Lambda_m$ is given by: </p> <p>$\Lambda_m = \frac{\kappa}{c}.$</p></p> <p><p>Substituting the values: </p> <p>$$\Lambda_m = \frac{1.15\; \text{S/m}}{50\; \text{mol/m}^3} = 0.023\; \text{S}\cdot \text{m}^2/\text{mol}.$$</p></p> <p>Convert the molar conductivity to the desired units (mS dm² mol⁻¹). </p> <p><p>First, convert the conductivity: </p> <p>$$0.023\; \text{S}\cdot \text{m}^2/\text{mol} \times 1000\; \frac{\text{mS}}{\text{S}} = 23\; \text{mS}\cdot \text{m}^2/\text{mol}.$$</p></p> <p><p>Now, convert the area units. Recall that: </p> <p>$1\; \text{m}^2 = 100\; \text{dm}^2.$ </p> <p>So, </p> <p>$$23\; \text{mS}\cdot \text{m}^2/\text{mol} = 23 \times 100\; \text{mS}\cdot \text{dm}^2/\text{mol} = 2300\; \text{mS}\cdot \text{dm}^2/\text{mol}.$$</p></p> <p><p>The problem asks for the answer in the form </p> <p>$\text{[blank]} \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}.$ </p> <p>Expressing 2300 mS dm²/mol in this form: </p> <p>$$2300\; \text{mS}\cdot \text{dm}^2/\text{mol} = 23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}.$$</p></p> <p>Thus, the molar conductivity is </p> <p>$\boxed{23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}}.$</p>