How will you prepare 50 mL of 70% alcohol solution? 15 mL ethyl alcohol in 50 mL c 35 ml. ethyl alcohol in 50 ml solution solution B. 25 ml ethyl alcohol in 50 ml D. 45 ml ethyl alcohol in 50 ml solution solution 85 grams of
The Correct Answer and Explanation is:
C. 35 mL ethyl alcohol in 50 mL solution
✅ Correct Answer: 35 mL ethyl alcohol in 50 mL solution
📘 Explanation
A 70% alcohol solution means that 70% of the final solution’s volume must be pure ethyl alcohol, and the remaining 30% must be water (or another suitable diluent).
You are asked to prepare 50 mL of this solution, so you need to calculate how much of that 50 mL should be alcohol:Volume of alcohol=70% of 50 mL=0.70×50=35 mL\text{Volume of alcohol} = 70\% \text{ of } 50 \text{ mL} = 0.70 \times 50 = 35 \text{ mL}Volume of alcohol=70% of 50 mL=0.70×50=35 mL
This means you need 35 mL of ethyl alcohol, and the rest (15 mL) will be water or diluent to reach a total volume of 50 mL.
Let’s now examine the options:
- A. 15 mL ethyl alcohol in 50 mL solution →
1550×100=30%\frac{15}{50} \times 100 = 30\%5015×100=30% → ❌ Too low - B. 25 mL ethyl alcohol in 50 mL solution →
2550×100=50%\frac{25}{50} \times 100 = 50\%5025×100=50% → ❌ Still too low - C. 35 mL ethyl alcohol in 50 mL solution →
3550×100=70%\frac{35}{50} \times 100 = 70\%5035×100=70% → ✅ Correct! - D. 45 mL ethyl alcohol in 50 mL solution →
4550×100=90%\frac{45}{50} \times 100 = 90\%5045×100=90% → ❌ Too high
🔬 Why It Matters:
Accurate preparation of alcohol solutions is essential, especially in medical, laboratory, or cleaning applications. Too little alcohol may reduce its effectiveness as a disinfectant, while too much could be unnecessarily harsh or unsafe.
Therefore, to prepare 50 mL of 70% alcohol, you need to combine:
- 35 mL of ethyl alcohol
- 15 mL of water or diluent
Final volume: 50 mL of 70% alcohol solution ✅
