The Molarity (M) of an aqueous solution containing $5.85 \mathrm{~g}$ of $\mathrm{NaCl}$ in $500 \mathrm{~mL}$ water is : (Given : Molar Mass $\mathrm{Na}: 23$ and $\mathrm{Cl}: 35.5 \mathrm{~gmol}^{-1}$)

<p>To find the molarity of the aqueous solution, we need to calculate the number of moles of NaCl and then divide it by the volume of the solution in liters.</p> <p>First, determine the molar mass of NaCl:</p> <p>$$ \text{Molar mass of NaCl} = \text{Molar mass of Na} + \text{Molar mass of Cl} = 23 + 35.5 = 58.5 \, \text{g/mol} $$</p> <p>Next, calculate the moles of NaCl:</p> <p>$$ \text{Moles of NaCl} = \frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} = 0.1 \, \text{mol} $$</p> <p>Now, convert the volume of the solution from milliliters to liters:</p> <p>$500 \, \text{mL} = 0.500 \, \text{L}$</p> <p>Finally, calculate the molarity (M), which is the number of moles of solute per liter of solution:</p> <p>$$ M = \frac{\text{Moles of NaCl}}{\text{Volume of solution in liters}} = \frac{0.1 \, \text{mol}}{0.500 \, \text{L}} = 0.2 \, \text{M} $$</p> <p>Therefore, the molarity of the solution is 0.2 M. The correct option is D.</p>