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 \( \text{Cl} = 35.5 \, \text{g mol}^{-1} \))
(Given: Molar Mass \( \text{Na} = 23 \) and \( \text{Cl} = 35.5 \, \text{g mol}^{-1} \))
- 20
- 0.2
- 2
- 4
The Correct Option is B
Solution and Explanation
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:
\[ M = \frac{n_{\text{NaCl}}}{V_{\text{sol}} \, (\text{in L})} = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{M} \]
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