Mixture Replacement: Formula, Rules and Solved Examples
Mixture Replacement deals with removing a fixed quantity from a mixture and replacing it with another liquid. The main method uses the fraction of mixture left after each replacement. This page covers the mixture replacement formula, repeated replacement, concentration calculations, reverse problems and concise solving tricks with numerical examples.
What Is Mixture Replacement?
The mixture must be uniform before every removal, so the removed quantity contains the same ratio of ingredients as the whole mixture. For replacement with water or another liquid containing no quantity of the considered substance, the amount of that substance after n operations is A(1 − x/V)^n, where A is its initial amount. The vessel volume remains V after each refill.
Mixture Replacement Formula & Tricks
Important Formulas
Here, V is the total vessel volume, x is the quantity replaced in each operation and n is the number of replacements.
A is the initial amount of the substance being measured. This formula applies when the replacing liquid contains none of that substance.
C_0 is the initial concentration and C_n is the concentration after n replacements.
C_r is the concentration of the liquid added at every replacement. For pure water, C_r = 0, giving C_n = C_0(1 − x/V)^n.
Quick Tricks
Instead of calculating the removed amount separately each time, find the fraction retained in one operation as (V − x)/V and raise it to the number of operations.
If the final amount and number of operations are given, divide by the retained fraction raised to n to find the initial amount.
After removing x from a volume V, the amount of the original substance left is its initial amount multiplied by (V − x)/V.
Mixture Replacement Concepts
Single Replacement of a Mixture
Because the mixture is uniform, removing x units removes x/V of every ingredient. Therefore, the fraction of each original ingredient left is 1 − x/V. If a 40 L mixture has 30 L milk and 10 L water, and 8 L is removed and replaced with water, milk left = 30 × 32/40 = 24 L.
Repeated Mixture Replacement
The same retained fraction applies after every operation because the vessel is refilled to V and mixed before the next removal. For a 60 L vessel, replacing 15 L each time leaves a fraction 45/60 = 3/4 after each operation. After 2 operations, the original substance remaining is A(3/4)^2 = 9A/16.
Concentration After Replacement with Water
If the initial concentration is C_0, the concentration after n replacements is C_0(1 − x/V)^n. For a 20% salt solution, replacing 10% of the vessel volume with water twice gives final concentration = 20% × (0.9)^2 = 16.2%.
Replacement with Another Mixture
The concentration difference from the replacing liquid is multiplied by the retained fraction in every operation. If C_r is zero, the formula becomes the usual water-replacement formula. For one replacement, this also gives C_1 = C_0(1 − x/V) + C_r(x/V).
Mixture Replacement Video Lessons
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Repeated Dilution Formula in Replacement
Learn the repeated dilution formula used in replacement method problems, including how successive removals and refills change the concentration or quantity of a mixture.
Practice Mixture Replacement Questions
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Mixture Replacement Revision Points
Use these formulas and rules for quick revision of replacement problems.
- The mixture must be thoroughly mixed before each quantity is removed.
- For vessel volume V and replacement quantity x, the retained fraction per operation is (V − x)/V.
- After n replacements, the original mixture fraction is (1 − x/V)^n.
- If the replacing liquid contains none of the considered substance, final amount = initial amount × (1 − x/V)^n.
- For replacement with water, final concentration = initial concentration × (1 − x/V)^n.
- If the added liquid has concentration C_r, use C_n = C_r + (C_0 − C_r)(1 − x/V)^n.
- The vessel volume is restored to V after every replacement.
Mixture Replacement FAQs
What is the mixture replacement formula?
If x is replaced from a vessel of volume V in every operation, the original mixture remaining after n operations is A(1 − x/V)^n, where A is the initial amount.
What fraction of a mixture remains after one replacement?
The fraction remaining is (V − x)/V, where V is the total volume and x is the quantity removed and replaced.
A 40 L mixture has 10 L replaced with water twice. What fraction of the original mixture remains?
The retained fraction per operation is 30/40 = 3/4. After two operations, the fraction remaining is (3/4)^2 = 9/16.
How much milk remains if 8 L is replaced from a 32 L vessel three times?
The retained fraction is 24/32 = 3/4. Therefore, milk remaining = initial milk × (3/4)^3 = 27/64 of the initial milk.
How is replacement with a liquid of different concentration calculated?
Use C_n = C_r + (C_0 − C_r)(1 − x/V)^n, where C_0 is the initial concentration and C_r is the concentration of the replacing liquid.
What happens when the entire mixture is removed and replaced?
If x = V, the retained fraction is zero. Therefore, none of the original mixture remains after one operation, and the vessel contains only the replacing liquid.