Question

100 mL of 2 M HCl is diluted to 500 mL. The molarity of the diluted solution is:

• A. 0.2 M
• B. 0.4 M
• C. 0.5 M
• D. 1.0 M

Hint:

Whenever a solution is diluted, the dilution factor is given by $\frac{V_2}{V_1}$.
Here, the volume is increased from $100\text{ mL}$ to $500\text{ mL}$ (a factor of $5$).
Thus, the concentration must decrease by a factor of $5$:
\[ M_2 = \frac{M_1}{5} = \frac{2}{5} = 0.4\text{ M} \]
This mental math technique helps save time in the exam.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question belongs to the topic "Solutions," specifically focusing on the dilution of solutions.
We are required to determine the final molarity of a hydrochloric acid (HCl) solution when its volume is increased from $100\text{ mL}$ to $500\text{ mL}$.

Step 2: Key Formula or Approach:
During dilution, the total amount (moles) of solute remains constant because only the solvent is added to the solution.
The relation between initial and final concentrations and volumes is given by the Dilution Equation:
\[ M_1 V_1 = M_2 V_2 \]
where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume.

Step 3: Detailed Explanation:

• We are given the following values:
Initial molarity ($M_1$) = $2\text{ M}$
Initial volume ($V_1$) = $100\text{ mL}$
Final volume ($V_2$) = $500\text{ mL}$

• We need to solve for the final molarity ($M_2$).

• Rearranging the dilution formula to isolate $M_2$:
\[ M_2 = \frac{M_1 V_1}{V_2} \]

• Substituting the values into the equation:
\[ M_2 = \frac{2\text{ M} \times 100\text{ mL}}{500\text{ mL}} \]

• Simplifying the calculations:
\[ M_2 = \frac{200}{500} = 0.4\text{ M} \]

• The process of dilution reduces the concentration of ions per unit volume, which lowers the molarity.



Step 4: Final Answer:
The final molarity of the diluted solution is $0.4\text{ M}$, which is option (B).