While making 100 ml of 2M NaCl solution, a student left the solution on a heating platform reducing the
volume by 50 ml. This solution was diluted 1:100 for use. What is the final concentration of NaCl in this
solution?
(1) 20 mM
(2) 80 mM
(3) 40 mM
(4) 400 mM
📘 Problem Statement
A student made 100 ml of 2M NaCl solution but left it on a heating platform, causing the volume to decrease by 50 ml. The solution was then diluted 1:100 for further use. What is the final concentration of NaCl in the diluted solution?
Options:
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20 mM
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80 mM
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40 mM
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400 mM
🔬 Steps for Solution Preparation and Concentration Calculation
To solve this problem, we will go step by step through the changes in the concentration of NaCl in the solution:
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Initial Concentration and Volume Change Due to Heating:
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The student initially prepared a 100 ml of 2M NaCl solution.
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The solution was heated, reducing its volume by 50 ml, leaving only 50 ml of the NaCl solution.
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As a result, the concentration of NaCl increases because the amount of NaCl stays the same, but the volume decreases.
New concentration after heating:
C1V1=C2V2C_1V_1 = C_2V_2
where:
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C1=2 MC_1 = 2 \, \text{M} (initial concentration),
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V1=100 mlV_1 = 100 \, \text{ml} (initial volume),
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V2=50 mlV_2 = 50 \, \text{ml} (final volume after heating),
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C2C_2 is the final concentration after the volume reduction.
Substituting the values:
2 M×100 ml=C2×50 ml2 \, \text{M} \times 100 \, \text{ml} = C_2 \times 50 \, \text{ml} C2=2 M×100 ml50 ml=4 MC_2 = \frac{2 \, \text{M} \times 100 \, \text{ml}}{50 \, \text{ml}} = 4 \, \text{M}
After heating, the concentration of the NaCl solution is 4M.
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Dilution Step:
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The solution is then diluted 1:100, meaning 1 part of the heated solution is mixed with 99 parts of solvent (usually water).
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The final concentration after dilution can be calculated using the dilution formula:
C1V1=C2V2C_1V_1 = C_2V_2
where:
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C1=4 MC_1 = 4 \, \text{M} (concentration after heating),
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V1=1 mlV_1 = 1 \, \text{ml} (amount taken for dilution),
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V2=100 mlV_2 = 100 \, \text{ml} (final volume after dilution),
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C2C_2 is the final concentration after dilution.
Substituting the values:
4 M×1 ml=C2×100 ml4 \, \text{M} \times 1 \, \text{ml} = C_2 \times 100 \, \text{ml} C2=4 M×1 ml100 ml=0.04 M=40 mMC_2 = \frac{4 \, \text{M} \times 1 \, \text{ml}}{100 \, \text{ml}} = 0.04 \, \text{M} = 40 \, \text{mM}
The final concentration of NaCl after dilution is 40 mM.
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✅ Correct Answer: (3) 40 mM
The final concentration of NaCl in the diluted solution is 40 mM.
💡 Why This Calculation is Important
This calculation illustrates the importance of understanding dilution and concentration in laboratory settings, especially when preparing solutions for experiments. Understanding how to adjust concentrations after heating or dilution is essential in many scientific fields, including:
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Biochemistry
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Pharmacology
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Environmental Science
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Chemical Engineering
✅ Key Takeaways
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Heating the NaCl solution reduces its volume, thereby increasing its concentration.
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Dilution involves mixing a concentrated solution with solvent, which decreases the concentration.
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Using the formula C1V1=C2V2C_1V_1 = C_2V_2, we can accurately calculate the new concentration after changes in volume and dilution.
