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

Answer: (B)

0.2

Answer 2

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:

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