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)

Given a 0.2 % (w/v) NaOH solution, 0.2 g of NaOH is present in 100 mL of solution.
Hence in $$1000\ \text{mL} = 1\ \text{dm}^3$$, mass of NaOH = $$0.2\times 10 = 2\ \text{g}$$.

Molar mass of NaOH = $$40\ \text{g mol}^{-1}$$.
Therefore molarity $$c$$ is
$$c = \frac{2\ \text{g}}{40\ \text{g mol}^{-1}} = 0.05\ \text{mol dm}^{-3}$$ $$-(1)$$

Resistivity of the solution, $$\rho = 870.0\ \text{m}\Omega\ \text{m}$$
$$1\ \text{m}\Omega = 10^{-3}\ \Omega$$, so
$$\rho = 870.0 \times 10^{-3}\ \Omega\ \text{m} = 0.870\ \Omega\ \text{m}$$ $$-(2)$$

Conductivity (specific conductance) $$\kappa$$ is the reciprocal of resistivity:
$$\kappa = \frac{1}{\rho} = \frac{1}{0.870} = 1.149\ \text{S m}^{-1}$$ $$-(3)$$

To work with molar conductivity we first convert $$\kappa$$ to $$\text{S cm}^{-1}$$:
$$1\ \text{S m}^{-1} = 0.01\ \text{S cm}^{-1}$$ (because $$1\ \text{m}=100\ \text{cm}$$ and conductivity $$\propto 1/{\text{length}}$$).
Hence
$$\kappa = 1.149 \times 0.01 = 0.01149\ \text{S cm}^{-1}$$ $$-(4)$$

Formula for molar conductivity $$\Lambda_m$$ (in $$\text{S cm}^2\ \text{mol}^{-1}$$):
$$\Lambda_m = \frac{1000\,\kappa}{c}$$ $$-(5)$$

Substituting $$\kappa$$ from $$(4)$$ and $$c$$ from $$(1)$$:
$$\Lambda_m = \frac{1000 \times 0.01149}{0.05} = 229.8\ \text{S cm}^2\ \text{mol}^{-1}$$ $$-(6)$$

Convert $$\Lambda_m$$ to the requested unit $$\text{mS dm}^2\ \text{mol}^{-1}$$.
Relation:
$$1\ \text{S cm}^2 = (1000\ \text{mS})(0.01\ \text{dm}^2) = 10\ \text{mS dm}^2$$.
Therefore
$$\Lambda_m = 229.8 \times 10 = 2298\ \text{mS dm}^2\ \text{mol}^{-1}$$ $$-(7)$$

Expressed as $$\Lambda_m = 22.98 \times 10^2\ \text{mS dm}^2\ \text{mol}^{-1}$$.
Nearest integer = $$23$$.

Hence the molar conductivity of the solution is 23 $$\times 10^2$$ mS dm$$^2$$ mol$$^{-1}$$.