Question

A solution is prepared by dissolving 5.85 g of NaCl (Molar mass = 58.5 g mol$^{-1}$) in water to make 500 mL of solution. The molarity of the solution is:

• A. 0.1 M
• B. 0.2 M
• C. 0.5 M
• D. 1.0 M

Hint:

If the volume is given in milliliters, you can use the combined formula for molarity to save time during the exam:
\[ M = \frac{\text{Weight of solute (g)} \times 1000}{\text{Molar mass of solute} \times \text{Volume of solution (mL)}} \]
Substituting: $M = \frac{5.85 \times 1000}{58.5 \times 500} = \frac{10 \times 1000}{100 \times 500} = 0.2\text{ M}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question belongs to the topic "Solutions," focusing on the concentration terms, specifically Molarity ($M$).
We are asked to calculate the molarity of a sodium chloride (NaCl) solution prepared by dissolving a given mass of solute in a specified volume of solution.

Step 2: Key Formula or Approach:
Molarity ($M$) is defined as the number of moles of solute dissolved per liter of solution:
\[ M = \frac{\text{Number of moles of solute } (n)}{\text{Volume of solution in liters } (V)} \]
The number of moles ($n$) is calculated by dividing the given mass ($w$) of the solute by its molar mass ($M_w$):
\[ n = \frac{w}{M_w} \]

Step 3: Detailed Explanation:

• We are given the following values:
Mass of NaCl ($w$) = $5.85\text{ g}$
Molar mass of NaCl ($M_w$) = $58.5\text{ g mol}^{-1}$
Volume of solution ($V$) = $500\text{ mL}$

• First, calculate the number of moles of NaCl solute:
\[ n = \frac{5.85\text{ g}}{58.5\text{ g mol}^{-1}} = 0.1\text{ mol} \]

• Next, convert the volume of the solution from milliliters to liters:
\[ V = \frac{500\text{ mL}}{1000\text{ mL/L}} = 0.5\text{ L} \]

• Finally, substitute the values of $n$ and $V$ into the molarity formula:
\[ M = \frac{0.1\text{ mol}}{0.5\text{ L}} = 0.2\text{ M} \]

• Molarity is a temperature-dependent property because the volume of the solution can expand or contract with temperature variations.



Step 4: Final Answer:
The molarity of the prepared NaCl solution is $0.2\text{ M}$, which corresponds to option (B).