The quantity of water (in ml) needed to reduce $9$ ml shaving lotion containing $50\%$ alcohol to a lotion containing $30\%$ alcohol, is

### Concept & Logic This is a classic dilution problem. By adding water, we are changing the total volume and the water content, but the absolute mass of the alcohol remains entirely unchanged. $$ \text{Initial Alcohol Mass} = \text{Final Alcohol Mass} $$ ### Step-by-Step Solution * **Given:** * Initial total volume = $9$ ml * Initial alcohol concentration = $50\%$ * Target alcohol concentration = $30\%$ * **Calculation:** * Find the absolute volume of alcohol in the original lotion: $50\%$ of $9$ ml $= 4.5$ ml. * Let the final total volume of the diluted lotion be $V$. * In the new lotion, the alcohol concentration must be $30\%$, but the absolute alcohol remains $4.5$ ml. * Set up the equation: $30\%$ of $V = 4.5$. * $0.30 \times V = 4.5$. * $V = \frac{4.5}{0.30} = 15$ ml. * The new total volume is $15$ ml. The initial volume was $9$ ml. * Quantity of water added = $15 - 9 = 6$ ml. ### Exam Strategy & Shortcut **Percentage-Volume Inversion:** The alcohol percentage drops from $50\%$ to $30\%$, which is a ratio of $5:3$. Because the alcohol volume is constant, the total mixture volume must be inversely proportional, which is a ratio of $3:5$. If the initial volume ($3$ units) $= 9$ ml, then $1$ unit $= 3$ ml. The final volume ($5$ units) $= 15$ ml. The added water is the difference ($2$ units) $= 2 \times 3 = 6$ ml. ### Common Pitfall Students frequently calculate the final total volume ($15$ ml) and accidentally select it as the answer if it's present in the options. Always reread the question to ensure you are answering what is asked—in this case, the *added* water, not the *final* volume. ### Final Answer **Therefore, the correct answer is 6.**