Question

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

• A. 20
• B. 0.2
• C. 2
• D. 4

Answer 1

The Correct Option is B

To calculate the molarity of the solution, first determine the molar mass of NaCl:

\[ \text{Molar mass of NaCl} = 23 \, \text{g/mol} + 35.5 \, \text{g/mol} = 58.5 \, \text{g/mol} \]

Calculate the number of moles of NaCl:

\[ n_{\text{NaCl}} = \frac{\text{Mass of NaCl}}{\text{Molar Mass of NaCl}} = \frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} = 0.1 \, \text{mol} \]

Given that the volume of the solution is 500 mL = 0.5 L, the molarity \( M \) is calculated as:

\[ M = \frac{n_{\text{NaCl}}}{V_{\text{sol}} \, (\text{in L})} = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]