ANAND CLASSES Study Material and Notes to learn how to use the molarity equation for mixing solutions of the same substance with step-by-step explanations, solved examples, MCQs, and conceptual clarity for JEE, NEET, and CBSE Class 11.
🧪 Concept: What Is Molarity?
Molarity (M) is the concentration of a solution expressed as: $$\text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$$
📘 When Do We Use the Mixing Formula?
When two solutions of the same solute but different molarities and volumes are mixed, the final concentration (molarity) changes depending on how much solute and solvent is present in each.
To find the final molarity after mixing, we use:
✅ Mixing Equation:
$$M_1V_1 + M_2V_2 = M_3V_3$$
📖 Meaning of Each Term:
| Symbol | Meaning |
|---|---|
| $M_1$ | Molarity of Solution 1 |
| $V_1$ | Volume of Solution 1 |
| $M_2$ | Molarity of Solution 2 |
| $V_2$ | Volume of Solution 2 |
| $M_3$ | Molarity of Final Solution |
| $V_3$ | Final Volume after mixing = $V_1 + V_2$ |
🔍 Derivation:
Let’s say:
- Solution 1 contains moles of solute = $M_1 \times V_1$
- Solution 2 contains moles of solute = $M_2 \times V_2$
When mixed together, the total moles of solute becomes: $$\text{Total moles} = M_1V_1 + M_2V_2$$
And the total volume is: $$V_3 = V_1 + V_2$$
Now, by definition of molarity: $$M_3 = \frac{\text{Total moles of solute}}{\text{Total volume}} = \frac{M_1V_1 + M_2V_2}{V_3}$$
Multiply both sides by $V_3$: $$M_1V_1 + M_2V_2 = M_3V_3$$
🧠 Important Notes:
- This equation works only if:
- The solute is the same in both solutions.
- There is no chemical reaction taking place.
- Volumes are additive (no volume contraction or expansion).
🎯 Solved Example:
Q. Mix 100 mL of 2 M NaCl with 300 mL of 1 M NaCl. What is the final molarity?
Solution:
Given:
- $M_1 = 2\,M,$ $V_1 = 100\,mL$
- $M_2 = 1\,M,$ $V_2 = 300\,mL$
Use: $$M_3 = \frac{M_1V_1 + M_2V_2}{V_1 + V_2} $$
$$M_3 = \frac{(2)(100) + (1)(300)}{100 + 300} = \frac{200 + 300}{400} = \frac{500}{400} = 1.25\,M$$
🧠 Quick Revision Points:
- Always convert volumes to same units.
- Formula assumes no reaction.
- Final volume is sum of individual volumes.
- You can rearrange the equation to find any missing variable.
🤔 Do You Know?
- This concept is useful in buffer preparations, dilutions, and laboratory mixing.
- If you add pure solvent (like water) to a solution, $M_2 = 0$
💡 Conceptual Questions (With Answers)
Q1. Two solutions of the same acid are mixed. One is more concentrated. Will the final molarity be higher or lower than both?
A: It will be in between the two original molarities.
Q2. Why can’t we use this equation for mixing NaOH and HCl?
A: Because a chemical reaction occurs (neutralization), so molarity changes due to new substances being formed.
📝 Worksheet Practice (With Answers)
Q1. 250 mL of 0.4 M HCl is mixed with 150 mL of 1.2 M HCl. What is the final molarity?
$$M_3 = \frac{(0.4)(250) + (1.2)(150)}{250 + 150} = \frac{100 + 180}{400} = \frac{280}{400} = 0.7\,M$$
Q2. 200 mL of 1 M NaOH is diluted with 800 mL water. What is the new molarity?
Since water has no solute, $M_2 = 0$
$$M_3 = \frac{(1)(200) + (0)(800)}{200 + 800} = \frac{200}{1000} = 0.2\,M$$
🧪 MCQs on Molarity and Mixing of Solutions
Q1.
100 mL of 2 M NaOH is mixed with 300 mL of 1 M NaOH. What is the final molarity of the mixture?
A. 1.25 M
B. 1.5 M
C. 1.75 M
D. 2.5 M
✅ Answer: A. 1.25 M
📘 Explanation: $$M_3 = \frac{M_1V_1 + M_2V_2}{V_1 + V_2} = \frac{(2)(100) + (1)(300)}{100 + 300} = \frac{500}{400} = 1.25\,M$$
Q2.
Which of the following conditions must be true to use the formula $$M_1V_1 + M_2V_2 = M_3V_3 $$
A. The solutions must contain the same solute
B. The solutions must be acidic
C. There must be a neutralization reaction
D. Volumes should always be in liters
✅ Answer: A. The solutions must contain the same solute
📘 Explanation: The mixing formula is based on additive moles and applies only when the solute is the same and no reaction occurs.
Q3.
If 250 mL of 1 M HCl is mixed with 250 mL of 3 M HCl, what is the molarity of the final solution?
A. 1.5 M
B. 2 M
C. 3 M
D. 4 M
✅ Answer: B. 2 M
📘 Explanation: $$M_3 = \frac{(1)(250) + (3)(250)}{250 + 250} = \frac{250 + 750}{500} = \frac{1000}{500} = 2\,M$$
Q4.
500 mL of 0.5 M NaCl solution is diluted by adding 500 mL of water. What is the final molarity?
A. 0.25 M
B. 0.50 M
C. 0.75 M
D. 1.00 M
✅ Answer: A. 0.25 M
📘 Explanation: $$M_3 = \frac{(0.5)(500) + (0)(500)}{500 + 500} = \frac{250}{1000} = 0.25\,M$$
Q5.
Which of the following is not a valid assumption in the use of the mixing formula?
A. No chemical reaction takes place
B. Solute is same in both solutions
C. Volumes are additive
D. Temperature must increase during mixing
✅ Answer: D. Temperature must increase during mixing
📘 Explanation: Temperature change is not required for using the formula. The other three are necessary assumptions.
