The Central Idea
If a mixture has total quantity \(Q\) and component concentration \(c\), then the amount of the component is
\[ \boxed{A=cQ}. \]Concentration changes when either the component amount \(A\), the total quantity \(Q\), or both change.
The fundamental concentration relationship is
\[ \boxed{c=\frac{A}{Q}}. \]Dilution and replacement problems are solved by tracking \(A\) and \(Q\) separately.
Three Basic Ways Concentration Can Change
Concentration can change by:
- adding material;
- removing material;
- removing and replacing material.
The key question is always:
How much of the component is present after the operation?
Dilution by Adding Pure Solvent
Suppose a solution contains \(A\) units of solute in total quantity \(Q\).
If \(x\) units of pure solvent are added, the solute amount remains
\[ A, \]while total quantity becomes
\[ Q+x. \]Therefore, the new concentration is
\[ \boxed{ c_{\text{new}}=\frac{A}{Q+x} }. \]Example: Dilute a Solution
A \(20\)-liter solution is \(40\%\) solute. How much water must be added to reduce the concentration to \(25\%\)?
Initial solute:
\[ A=0.40(20)=8. \]Let \(x\) liters of water be added.
Since no solute is added,
\[ 8=0.25(20+x). \]Solve:
\[ 8=5+0.25x. \] \[ 3=0.25x. \] \[ x=12. \]Therefore,
\[ \boxed{12\text{ liters of water}}. \]The Dilution Formula
When dilution changes only the total quantity and does not change the amount of solute, then
\[ c_1Q_1=c_2Q_2. \]Therefore,
\[ \boxed{ c_1Q_1=c_2Q_2 }. \]This formula is useful only when the component amount is unchanged between the initial and final states.
Example: Use the Dilution Formula
A \(15\)-liter solution is \(60\%\) solute. To what total volume must it be diluted to obtain a \(36\%\) solution?
Use
\[ 0.60(15)=0.36Q_2. \] \[ 9=0.36Q_2. \] \[ Q_2=25. \]The final total volume is
\[ \boxed{25\text{ liters}}. \]Since the original volume was \(15\) liters, the amount of solvent added is
\[ 25-15=10. \]Therefore,
\[ \boxed{10\text{ liters of solvent}}. \]Strengthening by Adding Pure Solute
Suppose a mixture initially contains component amount \(A\) in total quantity \(Q\).
Adding \(x\) units of pure component increases both the component amount and total quantity by \(x\):
\[ A_{\text{new}}=A+x, \] \[ Q_{\text{new}}=Q+x. \]Therefore,
\[ \boxed{ c_{\text{new}} = \frac{A+x}{Q+x} }. \]Example: Add Pure Solute
A \(30\)-kg mixture is \(20\%\) component A. How much pure component A must be added to raise the concentration to \(35\%\)?
Initial component amount:
\[ 0.20(30)=6. \]Let \(x\) kilograms of pure component be added.
\[ 6+x=0.35(30+x). \]Expand:
\[ 6+x=10.5+0.35x. \] \[ 0.65x=4.5. \] \[ x=\frac{90}{13}\approx6.923. \]Therefore,
\[ \boxed{6.92\text{ kg approximately}}. \]Removing Pure Component
If pure component is removed directly, then both the component amount and total quantity decrease by the removed amount.
If \(x\) units of pure component are removed,
\[ A_{\text{new}}=A-x, \] \[ Q_{\text{new}}=Q-x. \]Therefore,
\[ \boxed{ c_{\text{new}} = \frac{A-x}{Q-x} }. \]This model differs from removing a portion of a well-mixed solution.
Removing Part of a Well-Mixed Solution
If a solution is thoroughly mixed, every removed portion has the same concentration as the whole mixture at that moment.
Suppose the mixture has concentration \(c\), and \(x\) units are removed.
The removed component amount is
\[ cx. \]Therefore,
\[ A_{\text{new}}=A-cx. \]Removing a Well-Mixed Portion Does Not Change Concentration
Suppose a mixture has quantity \(Q\), component amount \(cQ\), and concentration \(c\).
Remove \(x\) units.
Remaining component:
\[ cQ-cx=c(Q-x). \]Remaining total quantity:
\[ Q-x. \]Therefore, the remaining concentration is
\[ \frac{c(Q-x)}{Q-x}=c. \]Thus,
\[ \boxed{\text{removing a well-mixed portion alone does not change concentration}}. \]Example: Remove Part of a Solution
A \(25\)-liter solution is \(40\%\) solute. Five liters are removed.
Solute removed:
\[ 0.40(5)=2. \]Initial solute:
\[ 0.40(25)=10. \]Remaining solute:
\[ 10-2=8. \]Remaining solution:
\[ 25-5=20. \]Therefore,
\[ \frac8{20}=0.40. \]The concentration remains
\[ \boxed{40\%}. \]Replacement Problems
A replacement problem removes part of a well-mixed mixture and then replaces the removed amount with another material.
A common case is:
- remove \(x\) units of mixture;
- replace them with \(x\) units of pure solvent.
The total quantity returns to its original value, but the component amount decreases.
One-Step Replacement with Pure Solvent
Begin with total quantity \(Q\) and concentration \(c\).
Remove \(x\) units.
The fraction of the mixture remaining is
\[ \frac{Q-x}{Q} = 1-\frac{x}{Q}. \]Therefore, the fraction of the original component remaining is also
\[ \boxed{ 1-\frac{x}{Q} }. \]One-Step Replacement Formula
The original component amount is
\[ cQ. \]After removing \(x\) units of well-mixed solution, the remaining component amount is
\[ cQ\left(1-\frac{x}{Q}\right). \]Replacing with pure solvent adds no component.
Since the total quantity returns to \(Q\), the new concentration is
\[ \boxed{ c_{\text{new}} = c\left(1-\frac{x}{Q}\right) }. \]Example: One Replacement
A \(20\)-liter solution is \(50\%\) solute. Four liters are removed and replaced with pure water.
The fraction retained is
\[ 1-\frac4{20} = \frac45. \]Therefore, the new concentration is
\[ 0.50\left(\frac45\right) = 0.40. \]Thus,
\[ \boxed{40\%}. \]The Same Example by Component Accounting
Initial solute:
\[ 0.50(20)=10. \]Four liters of \(50\%\) solution contain
\[ 0.50(4)=2 \]liters of solute.
Remaining solute:
\[ 10-2=8. \]Replacing with \(4\) liters of water restores the total to \(20\) liters.
Therefore,
\[ \frac8{20}=0.40. \]The two methods agree.
Repeated Replacement
If the same fraction of a mixture is removed and replaced repeatedly, the amount of the original component decreases geometrically.
If the retained fraction each time is
\[ r=1-\frac{x}{Q}, \]then after \(n\) identical replacements,
\[ \boxed{ c_n=c_0r^n }. \]Repeated Replacement Formula
For total quantity \(Q\), replacement amount \(x\), initial concentration \(c_0\), and \(n\) repeated replacements with pure solvent,
\[ \boxed{ c_n = c_0 \left( 1-\frac{x}{Q} \right)^n }. \]This formula assumes the mixture is thoroughly mixed before every removal.
Example: Two Replacements
A \(30\)-liter solution is \(60\%\) solute. Six liters are removed and replaced with water. The mixture is stirred, then the same process is repeated.
Retained fraction each time:
\[ 1-\frac6{30} = \frac45. \]After two replacements:
\[ c_2 = 0.60\left(\frac45\right)^2. \] \[ c_2 = 0.60\left(\frac{16}{25}\right) = 0.384. \]Therefore,
\[ \boxed{38.4\%}. \]Why Repeated Replacement Is Geometric
Each replacement removes the same fraction of whatever component is present at that moment.
The component amount is repeatedly multiplied by the same retained factor:
\[ A_1=A_0r, \] \[ A_2=A_1r=A_0r^2, \] \[ A_3=A_0r^3. \]Therefore,
\[ \boxed{A_n=A_0r^n}. \]Replacement with a Nonzero-Concentration Fluid
Replacement material need not be pure solvent.
Suppose a fraction
\[ f=\frac{x}{Q} \]of the mixture is removed and replaced with material having concentration \(c_r\).
The new concentration is
\[ \boxed{ c_{\text{new}} = c(1-f)+c_rf }. \]Deriving the General Replacement Formula
After removing fraction \(f\), fraction
\[ 1-f \]of the original mixture remains.
Its component contribution is
\[ c(1-f). \]The replacement contributes
\[ c_rf. \]Therefore,
\[ \boxed{ c_{\text{new}} = c(1-f)+c_rf }. \]Example: Replace with Another Solution
A \(20\)-liter solution is \(50\%\) solute. Four liters are removed and replaced with a \(10\%\) solution.
The replaced fraction is
\[ f=\frac4{20}=0.20. \]Therefore,
\[ c_{\text{new}} = 0.50(0.80)+0.10(0.20). \] \[ c_{\text{new}} = 0.40+0.02 = 0.42. \]Thus,
\[ \boxed{42\%}. \]Repeated Replacement toward a New Concentration
If each replacement uses a fluid with concentration \(c_r\), the mixture moves toward \(c_r\).
Let
\[ r=1-\frac{x}{Q}. \]Then after \(n\) replacements,
\[ \boxed{ c_n = c_r+(c_0-c_r)r^n }. \]When \(c_r=0\), this reduces to the pure-solvent formula.
Why the General Repeated Formula Works
The difference between the current concentration and the replacement concentration is multiplied by the retained fraction each cycle.
Thus,
\[ c_{n+1}-c_r = r(c_n-c_r). \]Repeating gives
\[ c_n-c_r = (c_0-c_r)r^n. \]Therefore,
\[ c_n = c_r+(c_0-c_r)r^n. \]Unknown Replacement Amount
Suppose one replacement with pure solvent changes concentration from \(c_0\) to \(c_1\) in a container of volume \(Q\).
Then
\[ c_1 = c_0\left(1-\frac{x}{Q}\right). \]Solve for \(x\):
\[ \frac{c_1}{c_0} = 1-\frac{x}{Q}. \] \[ \frac{x}{Q} = 1-\frac{c_1}{c_0}. \]Therefore,
\[ \boxed{ x = Q\left(1-\frac{c_1}{c_0}\right) }. \]Example: Find the Replacement Amount
A \(40\)-liter solution is \(50\%\) solute. How much must be removed and replaced with pure water to reduce the concentration to \(35\%\) in one step?
Use
\[ 0.35 = 0.50\left(1-\frac{x}{40}\right). \]Divide by \(0.50\):
\[ 0.70 = 1-\frac{x}{40}. \] \[ \frac{x}{40}=0.30. \] \[ x=12. \]Therefore,
\[ \boxed{12\text{ liters}}. \]Unknown Number of Replacements
If repeated pure-solvent replacement gives
\[ c_n = c_0r^n, \]then the number of replacement cycles can be found from
\[ \frac{c_n}{c_0}=r^n. \]Taking logarithms,
\[ \boxed{ n = \frac{\ln(c_n/c_0)}{\ln r} }. \]The final practical answer may need to be rounded up to the next whole replacement cycle.
Example: How Many Replacements?
A \(20\)-liter solution begins at \(80\%\) concentration. Each cycle removes \(5\) liters and replaces them with water. How many complete cycles are needed to reduce the concentration below \(30\%\)?
The retained fraction is
\[ r=1-\frac5{20}=\frac34. \]We want
\[ 0.80\left(\frac34\right)^n<0.30. \]Equivalently,
\[ \left(\frac34\right)^n<0.375. \]Testing powers:
\[ \left(\frac34\right)^3\approx0.4219, \] \[ \left(\frac34\right)^4\approx0.3164. \]Therefore,
\[ 0.80\left(\frac34\right)^4\approx0.2531<0.30. \]Thus,
\[ \boxed{4\text{ replacement cycles}}. \]Replacement versus Simple Dilution
Dilution by adding solvent increases total quantity.
Replacement removes mixture first and then restores the original total quantity.
These are different operations.
In dilution,
\[ Q_{\text{final}}>Q_{\text{initial}}. \]In equal-volume replacement,
\[ Q_{\text{final}}=Q_{\text{initial}}. \]Replacement versus Removing Pure Solute
Removing a well-mixed portion removes all components proportionally.
Removing pure solute selectively removes only the component.
The algebra and final concentration differ greatly.
Always identify what exactly is being removed.
Sequential Operations
Some problems involve several different operations.
For example:
- remove part of a mixture;
- add pure solvent;
- add concentrated solution;
- remove another portion.
Update the component amount and total quantity after every step.
Example: Sequential Operations
A \(20\)-liter solution is \(40\%\) solute. Remove \(5\) liters, then add \(8\) liters of pure water.
Initial solute:
\[ 0.40(20)=8. \]Removed solute:
\[ 0.40(5)=2. \]Remaining solute:
\[ 8-2=6. \]Remaining solution after removal:
\[ 15\text{ L}. \]Add \(8\) liters of water:
\[ Q_{\text{final}}=23. \]Therefore,
\[ c_f=\frac6{23}\approx0.2609. \]Thus,
\[ \boxed{26.09\%\text{ approximately}}. \]Unknown Added Solvent after Removal
Suppose a mixture is partially removed and then an unknown amount of pure solvent is added.
First compute the remaining component amount after removal.
Then let the final quantity be
\[ Q_{\text{remaining}}+x. \]The target concentration equation becomes
\[ \boxed{ A_{\text{remaining}} = c_f(Q_{\text{remaining}}+x) }. \]Replacement Tables
A replacement table can track the state before and after each step.
\[ \begin{array}{c|c|c|c} \text{stage} & \text{total quantity} & \text{concentration} & \text{component}\\ \hline \text{initial} & Q & c_0 & c_0Q\\ \text{after removal} & Q-x & c_0 & c_0(Q-x)\\ \text{after replacement} & Q & c_1 & c_1Q \end{array} \]This layout is especially useful before introducing a shortcut formula.
Repeated Replacement as Exponential Decay
The repeated replacement model
\[ c_n=c_0r^n \]is a discrete exponential decay model when
\[ 0<r<1. \]Each replacement cycle multiplies the current concentration by the same retained fraction.
This connects mixture problems to exponential models studied elsewhere in algebra.
Limiting Behavior
For repeated replacement with pure solvent,
\[ c_n=c_0r^n, \qquad 0<r<1. \]As the number of replacements grows,
\[ r^n\to0. \]Therefore,
\[ \boxed{c_n\to0}. \]The concentration approaches zero, although a finite number of replacement cycles does not generally remove every last trace.
Replacement toward a Nonzero Limit
If replacement fluid has concentration \(c_r\), then
\[ c_n = c_r+(c_0-c_r)r^n. \]Since
\[ r^n\to0, \]we obtain
\[ \boxed{c_n\to c_r}. \]The mixture approaches the concentration of the replacement fluid.
Checking Dilution Problems
Adding pure solvent should lower concentration.
The component amount should remain unchanged.
Therefore, verify both:
\[ A_{\text{before}}=A_{\text{after}} \]and
\[ c_{\text{after}}<c_{\text{before}}. \]Checking Pure-Solute Addition
Adding pure component should increase concentration toward \(100\%\).
The final concentration should satisfy
\[ c_{\text{initial}} < c_{\text{final}} < 1 \]for a positive finite addition to a non-pure initial mixture.
Checking Replacement Problems
After replacement with pure solvent, the total quantity should be restored to its original value.
The component amount should decrease by exactly the component removed.
The final concentration should be lower than the original concentration.
Checking Repeated Replacement
If each cycle retains fraction \(r\), then after \(n\) cycles the remaining original component fraction should be
\[ r^n. \]Because
\[ 0<r<1, \]the concentration should decrease monotonically toward the replacement concentration.
Common Error: Using the Dilution Formula When Solute Changes
The formula
\[ c_1Q_1=c_2Q_2 \]assumes the component amount is unchanged.
It cannot be used when pure solute is added, when mixture is removed, or when the component amount otherwise changes.
Common Error: Treating Removed Mixture as Pure Solvent
A removed portion of a well-mixed solution contains solute in the same proportion as the original mixture.
Removing \(x\) liters from a \(30\%\) solution removes
\[ \boxed{0.30x} \]liters of solute.
Common Error: Assuming Removal Alone Changes Concentration
Removing part of a well-mixed solution reduces both solute and total solution proportionally.
Therefore, removal alone leaves concentration unchanged.
Common Error: Replacing before Mixing
Repeated replacement formulas assume the mixture is thoroughly mixed before each removal.
Without remixing, the removed portion may not have the same concentration as the entire container.
The standard formula would then not apply.
Common Error: Subtracting the Same Absolute Component Amount Repeatedly
In repeated replacement, each cycle removes the same fraction of the component currently present, not the same absolute component amount.
That is why the process is multiplicative:
\[ \boxed{A_n=A_0r^n}. \]Common Error: Forgetting the Total Quantity after Addition
If \(x\) liters are added to \(Q\) liters, the final quantity is
\[ \boxed{Q+x}. \]Using the original denominator \(Q\) after addition produces an incorrect concentration.
Common Error: Forgetting the Total Quantity after Removal
If \(x\) units are removed from a mixture of quantity \(Q\), the remaining total is
\[ \boxed{Q-x}. \]The denominator must reflect the new total unless the removed quantity is replaced.
Common Error: Using Percentage Points as Multipliers
A concentration change from \(40\%\) to \(30\%\) is a decrease of \(10\) percentage points.
But the retained concentration factor is
\[ \frac{30}{40}=0.75, \]not \(0.90\).
Distinguish percentage-point differences from multiplicative percent changes.
A Reliable Procedure for Dilution Problems
- Compute the initial component amount \(A=cQ\).
- Identify what is being added.
- If pure solvent is added, keep \(A\) unchanged.
- Write the new total quantity.
- Use \(c=A/Q\) or \(c_1Q_1=c_2Q_2\) when valid.
- Solve for the unknown addition or final quantity.
- Check that concentration moves in the expected direction.
A Reliable Procedure for Removal Problems
- Assume the mixture is well mixed unless stated otherwise.
- Compute the current concentration.
- Multiply concentration by the removed quantity to find component removed.
- Subtract both total quantity and component amount.
- Recompute concentration if further operations follow.
- Remember that removal alone does not change concentration.
A Reliable Procedure for Replacement Problems
- Identify the total quantity \(Q\).
- Identify the removed and replaced quantity \(x\).
- Compute the retained fraction \(1-x/Q\).
- Multiply the current component amount or concentration by that retained fraction.
- Add the contribution of the replacement fluid if its concentration is nonzero.
- Repeat the process or use a power formula for identical repeated cycles.
- Verify that total quantity returns to \(Q\) after each equal replacement.
Practice 1: Simple Dilution
A \(24\)-liter solution is \(50\%\) solute. How much water must be added to reduce the concentration to \(30\%\)?
Practice 2: Final Volume after Dilution
Ten liters of an \(80\%\) solution are diluted to \(40\%\). Find the final total volume.
Practice 3: Add Pure Solute
A \(20\)-kg mixture is \(25\%\) component A. How much pure component A must be added to raise the concentration to \(40\%\)?
Practice 4: Remove Well-Mixed Solution
A \(30\)-liter solution is \(35\%\) solute. Six liters are removed. Find the remaining total amount, remaining solute amount, and concentration.
Practice 5: One-Step Replacement
A \(25\)-liter solution is \(60\%\) solute. Five liters are removed and replaced with water. Find the new concentration.
Practice 6: Two Replacements
A \(40\)-liter solution is \(50\%\) solute. Eight liters are removed and replaced with water, the mixture is stirred, and the process is repeated. Find the final concentration.
Practice 7: Three Replacements
A \(30\)-liter solution is \(90\%\) solute. Three liters are removed and replaced with water three times, mixing before each removal. Find the final concentration.
Practice 8: Replace with Another Solution
A \(20\)-liter mixture is \(50\%\) solute. Four liters are removed and replaced with a \(25\%\) solution. Find the new concentration.
Practice 9: Find Replacement Amount
A \(50\)-liter solution is \(40\%\) solute. How much must be removed and replaced with water to reduce the concentration to \(28\%\) in one step?
Practice 10: Sequential Removal and Dilution
A \(20\)-liter solution is \(45\%\) solute. Four liters are removed, then \(10\) liters of water are added. Find the final concentration.
Practice 11: Unknown Added Water after Removal
A \(30\)-liter solution is \(40\%\) solute. Six liters are removed. How much water must then be added so the final concentration is \(25\%\)?
Practice 12: Repeated Replacement Threshold
A \(20\)-liter solution is \(80\%\) solute. Each cycle removes \(4\) liters and replaces them with water. How many complete cycles are needed to reduce the concentration below \(35\%\)?
Practice 13: Nonzero Replacement Limit
A mixture is initially \(70\%\) solute. Each cycle replaces \(20\%\) of the mixture with a \(10\%\) solution. Find the concentration after two cycles.
Practice 14: Identify the Invariant
A solution is diluted only by adding pure water. Which quantity remains constant: total volume, solute amount, or concentration?
Practice 15: Identify the Error
A \(40\%\) solution has \(10\) liters removed. A student says the remaining concentration must be less than \(40\%\) because some solute was removed.
Explain the error.
Practice Answers
Practice 1
Initial solute:
\[ 0.50(24)=12. \]Let \(x\) liters of water be added:
\[ 12=0.30(24+x). \] \[ 12=7.2+0.3x. \] \[ 4.8=0.3x. \] \[ x=16. \]Therefore,
\[ \boxed{16\text{ liters}}. \]Practice 2
\[ 0.80(10)=0.40Q_2. \] \[ 8=0.40Q_2. \] \[ Q_2=20. \]Therefore,
\[ \boxed{20\text{ liters}}. \]Practice 3
Initial component:
\[ 0.25(20)=5. \]Let \(x\) kilograms of pure component be added:
\[ 5+x=0.40(20+x). \] \[ 5+x=8+0.4x. \] \[ 0.6x=3. \] \[ x=5. \]Therefore,
\[ \boxed{5\text{ kg}}. \]Practice 4
Initial solute:
\[ 0.35(30)=10.5. \]Removed solute:
\[ 0.35(6)=2.1. \]Remaining solute:
\[ 10.5-2.1=8.4. \]Remaining total:
\[ 30-6=24. \]Concentration:
\[ \frac{8.4}{24}=0.35. \]Therefore,
\[ \boxed{24\text{ L},\quad8.4\text{ L solute},\quad35\%}. \]Practice 5
Retained fraction:
\[ 1-\frac5{25}=0.80. \]New concentration:
\[ 0.60(0.80)=0.48. \]Therefore,
\[ \boxed{48\%}. \]Practice 6
Retained fraction each cycle:
\[ 1-\frac8{40}=0.80. \]After two cycles:
\[ c_2=0.50(0.80)^2. \] \[ c_2=0.50(0.64)=0.32. \]Therefore,
\[ \boxed{32\%}. \]Practice 7
Retained fraction:
\[ 1-\frac3{30}=0.90. \]After three cycles:
\[ c_3=0.90(0.90)^3. \] \[ c_3=0.90(0.729)=0.6561. \]Therefore,
\[ \boxed{65.61\%}. \]Practice 8
Replaced fraction:
\[ \frac4{20}=0.20. \]Therefore,
\[ c_{\text{new}} = 0.50(0.80)+0.25(0.20). \] \[ c_{\text{new}} = 0.40+0.05 = 0.45. \]Thus,
\[ \boxed{45\%}. \]Practice 9
\[ 0.28 = 0.40\left(1-\frac{x}{50}\right). \] \[ 0.70 = 1-\frac{x}{50}. \] \[ \frac{x}{50}=0.30. \] \[ x=15. \]Therefore,
\[ \boxed{15\text{ liters}}. \]Practice 10
Initial solute:
\[ 0.45(20)=9. \]Removed solute:
\[ 0.45(4)=1.8. \]Remaining solute:
\[ 7.2. \]Remaining volume after removal:
\[ 16. \]Add \(10\) liters water:
\[ Q_f=26. \]Therefore,
\[ c_f=\frac{7.2}{26}\approx0.2769. \]Thus,
\[ \boxed{27.69\%\text{ approximately}}. \]Practice 11
Initial solute:
\[ 0.40(30)=12. \]Removed solute:
\[ 0.40(6)=2.4. \]Remaining solute:
\[ 9.6. \]Remaining volume:
\[ 24. \]Let \(x\) liters of water be added:
\[ 9.6=0.25(24+x). \] \[ 9.6=6+0.25x. \] \[ 3.6=0.25x. \] \[ x=14.4. \]Therefore,
\[ \boxed{14.4\text{ liters}}. \]Practice 12
Retained fraction:
\[ 1-\frac4{20}=0.80. \]We need
\[ 0.80(0.80)^n<0.35. \]Test:
\[ n=3: \qquad 0.80(0.80)^3=0.4096. \] \[ n=4: \qquad 0.80(0.80)^4=0.32768. \]Therefore,
\[ \boxed{4\text{ cycles}}. \]Practice 13
Here
\[ c_0=0.70, \qquad c_r=0.10, \qquad r=0.80. \]After two cycles:
\[ c_2 = 0.10+(0.70-0.10)(0.80)^2. \] \[ c_2 = 0.10+0.60(0.64). \] \[ c_2=0.484. \]Therefore,
\[ \boxed{48.4\%}. \]Practice 14
Adding pure water changes total volume and concentration but does not change the amount of solute.
Therefore,
\[ \boxed{\text{solute amount remains constant}}. \]Practice 15
Removing part of a well-mixed solution removes solute and solvent in the same proportion as they occur in the mixture.
Therefore, removal alone does not change concentration.
\[ \boxed{\text{The remaining solution is still }40\%}. \]What to Remember
Concentration is
\[ \boxed{c=\frac{A}{Q}}. \]Dilution by pure solvent keeps the component amount constant:
\[ \boxed{c_1Q_1=c_2Q_2}. \]Removing a well-mixed portion alone does not change concentration.
For one replacement with pure solvent,
\[ \boxed{ c_{\text{new}} = c_0\left(1-\frac{x}{Q}\right) }. \]For \(n\) identical replacements,
\[ \boxed{ c_n = c_0 \left( 1-\frac{x}{Q} \right)^n }. \]The most important habit is to track component amount and total quantity separately after every operation.
Where We Go Next
Concentration problems focus on how much of a component is present in a mixture. The same conservation structure applies when the component is a metal in an alloy, a grade in a blended material, or a monetary value attached to each unit.
The next lesson, Alloy, Composition, Price, and Value Mixtures, extends the mixture framework to those applications.