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An analytical chemist standardizes an automated titrator using two delivery flow rates measured in millimoles per minute (mmol/min\text{mmol/min}) and micromoles per second (μmol/s\mu\text{mol/s}). The cumulative delivery equation after tt minutes is expressed as 150 t mmol+0.050 mol=2.50×103 μmol/s⋅(60 t s/min)+5.0×104 μmol150\, t\text{ mmol} + 0.050\text{ mol} = 2.50\times 10^3\,\mu\text{mol/s} \cdot (60\, t\text{ s/min}) + 5.0\times 10^4\,\mu\text{mol}. Determine whether there is a unique operational time tt, no time, or an infinite number of operating times tt at which both sides deliver equivalent molar amounts.

Answer and steps

Final answer

Infinitely many solutions for all t≥0t \ge 0. (Converting all terms to millimoles yields the identity 150t+50=150t+50150t + 50 = 150t + 50.)

Steps

01

Recall the relevant unit conversion factors

To determine whether the equation has a unique solution, no solution, or infinitely many solutions, simplify both sides into a common unit of measurement. A convenient unit for all terms is millimoles (mmol\text{mmol}).

  • 1 mol=103 mmol=1,000 mmol1\text{ mol} = 10^3\text{ mmol} = 1{,}000\text{ mmol}
  • 1 mmol=103 μmol=1,000 μmol1\text{ mmol} = 10^3\,\mu\text{mol} = 1{,}000\,\mu\text{mol}, which means 1 μmol=10−3 mmol=11,000 mmol1\,\mu\text{mol} = 10^{-3}\text{ mmol} = \frac{1}{1{,}000}\text{ mmol}
02

Convert the left-hand side (LHS) to millimoles

The left-hand side is: LHS=150t mmol+0.050 mol\text{LHS} = 150t\text{ mmol} + 0.050\text{ mol}

Convert the constant term from moles to millimoles: 0.050 mol=0.050×1,000 mmol=50 mmol0.050\text{ mol} = 0.050 \times 1{,}000\text{ mmol} = 50\text{ mmol}

Thus, in millimoles, the LHS becomes: LHS=150t+50\text{LHS} = 150t + 50

03

Convert the right-hand side (RHS) to millimoles

The right-hand side is: RHS=2.50×103 μmol/s⋅(60t s)+5.0×104 μmol\text{RHS} = 2.50\times 10^3\,\mu\text{mol/s} \cdot (60t\text{ s}) + 5.0\times 10^4\,\mu\text{mol}

  1. Simplify the time-dependent term: (2.50×103)⋅60t=150,000t μmol=1.50×105t μmol(2.50\times 10^3) \cdot 60t = 150{,}000t\,\mu\text{mol} = 1.50\times 10^5t\,\mu\text{mol} Convert to millimoles: 150,000t μmol1,000 μmol/mmol=150t mmol\frac{150{,}000t\,\mu\text{mol}}{1{,}000\,\mu\text{mol/mmol}} = 150t\text{ mmol}
  2. Convert the constant term to millimoles: 5.0×104 μmol=50,000 μmol5.0\times 10^4\,\mu\text{mol} = 50{,}000\,\mu\text{mol} 50,000 μmol1,000 μmol/mmol=50 mmol\frac{50{,}000\,\mu\text{mol}}{1{,}000\,\mu\text{mol/mmol}} = 50\text{ mmol}

Thus, in millimoles, the RHS becomes: RHS=150t+50\text{RHS} = 150t + 50

04

Compare both sides and interpret the result

Equating the simplified expressions: 150t+50=150t+50150t + 50 = 150t + 50

Subtract 150t150t from both sides: 50=5050 = 50

This produces a true mathematical statement that is independent of tt. An equation in the form a⋅t+b=a⋅t+ba \cdot t + b = a \cdot t + b is an identity.

Therefore, the equation is satisfied for any operational time t≥0t \ge 0, meaning there are infinitely many solutions.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 A student omits the conversion from seconds to minutes when multiplying the right-hand flow rate, concluding there is a unique operational time t = 0.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    LHS\text{LHS} = 150t + 0.050 ×\times 1000 = 150t + 50 mmol\text{ mmol}

  2. Step 2

    RHS\text{RHS} = (2.50×1031000)\left(\frac{2.50 \times 10^3}{1000}\right)t + (5.0×1041000)\left(\frac{5.0 \times 10^4}{1000}\right) = 2.5t + 50 mmol\text{ mmol}

    Correct

    RHS\text{RHS} = (2.50×103×601000)\left(\frac{2.50 \times 10^3 \times 60}{1000}\right)t + (5.0×1041000)\left(\frac{5.0 \times 10^4}{1000}\right) = 150t + 50 mmol\text{ mmol}

    You converted the units from micromoles to millimoles by dividing by 10001000, but you overlooked the time unit conversion factor of 60 s/min60\text{ s/min} explicitly given in the problem as (60t s/min)(60t\text{ s/min}). Because the flow rate is in μmol/s\mu\text{mol/s} and tt is measured in minutes, each minute contains 60 seconds60\text{ seconds}, making the rate 2.50×103×60/1000=150 mmol/min2.50 \times 10^3 \times 60 / 1000 = 150\text{ mmol/min}.

  3. Step 3

    150t + 50 = 2.5t + 50   ⟹  \implies 147.5t = 0   ⟹  \implies t = 0

    Correct

    150t + 50 = 150t + 50   ⟹  \implies 50 = 50

    You solved your linear equation algebraically correctly, but the coefficients were already distorted by the missing factor of 6060.

  4. Step 4

    There is a unique operational time \text{There is a unique operational time } t = 0 min\text{ min}.

    Correct

    Infinitely many solutions for all \text{Infinitely many solutions for all } t ≥\ge 0.

    Because the true equation simplifies to the identity 150t+50=150t+50150t + 50 = 150t + 50, both flow setups deliver equal molar amounts at all operating times t≥0t \ge 0, not just at t=0t = 0.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly recognized that all terms must be brought into a single common unit before solving.

You used the correct metric conversion factors 1 mol=1000 mmol1\text{ mol} = 1000\text{ mmol} and 1 mmol=1000 μmol1\text{ mmol} = 1000\,\mu\text{mol}.

You performed standard algebraic balancing accurately on the equation you formulated.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omitted Unit Conversion Factor

You dropped the factor of 6060 when multiplying the rate in μmol/s\mu\text{mol/s} by time tt in minutes.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You focused exclusively on the metric prefix conversion (micromoles to millimoles) and missed matching the time dimension units (seconds versus minutes).

REASONING PATH
Your ACEpath teacher’s suggested approach
Dual-Unit Verification

  • Compound rates with different base units (e.g., μmol/s\mu\text{mol/s} vs. mmol/min\text{mmol/min}) require converting both the quantity prefix and the time unit.

  • Always write out rate units explicitly with dimensional analysis: (2.50×103 μmols)⋅(1 mmol1000 μmol)⋅(60 s1 min)=150 mmolmin\left(2.50 \times 10^3\,\frac{\mu\text{mol}}{\text{s}}\right) \cdot \left(\frac{1\text{ mmol}}{1000\,\mu\text{mol}}\right) \cdot \left(\frac{60\text{ s}}{1\text{ min}}\right) = 150\,\frac{\text{mmol}}{\text{min}}.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Include the time conversion of 60 s/min60\text{ s/min}: RHS=2.50×103×601000 t+5.0×1041000=150t+50\text{RHS} = \frac{2.50 \times 10^3 \times 60}{1000}\,t + \frac{5.0 \times 10^4}{1000} = 150t + 50.

  2. Equate the two simplified sides: 150t+50=150t+50150t + 50 = 150t + 50.

  3. Subtract 150t150t from both sides to obtain 50=5050 = 50, which is an identity valid for all valid times.

Final answer

Infinitely many solutions for all t≥0t \ge 0.

NEXT STEPS
Final Takeaways from Your Teacher

  • Track units as explicit algebraic factors in every rate calculation.

  • Check whether linear equations simplify to an identity (0=00 = 0) or a contradiction (a=ba = b with a≠ba \ne b) once all units are unified.

  • Practice multi-step dimensional analysis problems that involve both prefix scaling and time unit changes.

OverviewACEpath Teacher's Diagnosis

02 A student incorrectly converts moles to millimoles by dividing by 1000 instead of multiplying, leading to the false conclusion that no operational time exists.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    LHS=150t+0.0501000=150t+0.00005 mmol\text{LHS} = 150t + \frac{0.050}{1000} = 150t + 0.00005\text{ mmol}

    Correct

    LHS=150t+(0.050×1000)=150t+50 mmol\text{LHS} = 150t + (0.050 \times 1000) = 150t + 50\text{ mmol}

    To convert from a larger base unit (moles) to a smaller milli-unit (millimoles), you must multiply by 1,0001{,}000, because 1 mol=1,000 mmol1\text{ mol} = 1{,}000\text{ mmol}. Dividing by 1,0001{,}000 resulted in 0.00005 mmol0.00005\text{ mmol} instead of 50 mmol50\text{ mmol}.

  2. Step 2

    RHS=(2.50×103×601000)t+(5.0×1041000)=150t+50 mmol\text{RHS} = \left(\frac{2.50\times 10^3 \times 60}{1000}\right)t + \left(\frac{5.0\times 10^4}{1000}\right) = 150t + 50\text{ mmol}

  3. Step 3

    150t+0.00005=150t+50  ⟹  0.00005=50 (impossible)150t + 0.00005 = 150t + 50 \implies 0.00005 = 50\text{ (impossible)}

    Correct

    150t+50=150t+50  ⟹  50=50150t + 50 = 150t + 50 \implies 50 = 50

    Because of the earlier conversion error on the left-hand side, the variable terms canceled to leave unequal constants (0.00005=500.00005 = 50). With the correct conversion, both sides are identical (50=5050 = 50), which is always true.

  4. Step 4

    There is no operational time t.\text{There is no operational time } t.

    Correct

    Infinitely many solutions for all t≥0\text{Infinitely many solutions for all } t \ge 0

    Since both sides reduce to the exact same linear expression 150t+50150t + 50, the equation is an identity, meaning there are infinitely many valid operating times t≥0t \ge 0.

WHAT YOU DID WELL
Compliments from Your Teacher

You converted units of time by multiplying seconds by 6060 to align with minutes.

You correctly divided micromoles by 1,0001{,}000 to convert them into millimoles on the right-hand side.

You correctly understood that an equation reducing to a false equality of constants indicates no solution.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Unit Conversion Direction Error

Dividing 0.050 mol0.050\text{ mol} by 1,0001{,}000 rather than multiplying by 1,0001{,}000 when converting to millimoles.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusion regarding when to multiply versus divide during metric prefix conversions. Converting from a larger unit to a smaller unit (extmol→mmol ext{mol} \to \text{mmol}) yields a numerically larger value, requiring multiplication by the conversion factor.

REASONING PATH
Your ACEpath teacher’s suggested approach
Larger to Smaller Unit Check

  • 1 mol=1,000 mmol1\text{ mol} = 1{,}000\text{ mmol}. Moving to a smaller unit increases the numerical magnitude.

  • Always multiply when converting from a base unit to a milli-prefixed unit: mol×1,000=mmol\text{mol} \times 1{,}000 = \text{mmol}.

Dimensional Analysis Setup

  • Multiply by a unit conversion factor equal to 11, arranging units so the unwanted unit cancels out.

  • Write 0.050 mol×1,000 mmol1 mol=50 mmol0.050\text{ mol} \times \frac{1{,}000\text{ mmol}}{1\text{ mol}} = 50\text{ mmol} to guarantee the correct arithmetic operation.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Convert the LHS constant term: 0.050 mol×1,000 mmol/mol=50 mmol0.050\text{ mol} \times 1{,}000\text{ mmol/mol} = 50\text{ mmol}.

  2. Write the LHS expression: LHS=150t+50 mmol\text{LHS} = 150t + 50\text{ mmol}.

  3. Equate LHS and RHS: 150t+50=150t+50150t + 50 = 150t + 50.

  4. Subtract 150t150t from both sides to get the identity: 50=5050 = 50.

Final answer

Infinitely many solutions for all t≥0t \ge 0.

NEXT STEPS
Final Takeaways from Your Teacher

  • Use dimensional analysis fractions explicitly so unit labels cancel out visibly before doing the arithmetic.

  • Perform a sanity check on unit conversions: smaller units must result in larger numerical counts.

  • Re-evaluate linear identities after fixing arithmetic to confirm whether coefficients and constants match.

Lock it in

Question

How do you convert a rate given in μmol/s\mu\text{mol/s} to mmol/min\text{mmol/min}?

Answer

Multiply by 601000\frac{60}{1000} (or 0.060.06). You divide by 10001000 for prefix change (μmol→mmol\mu\text{mol} \to \text{mmol}) and multiply by 6060 for time (seconds→minutes\text{seconds} \to \text{minutes}).

Similar question

01

A flow pump delivers solution at 40 mL/min+200 mL40\text{ mL/min} + 200\text{ mL}, while a second pump delivers at 0.5 mL/s⋅(60t s/min)+0.2 L0.5\text{ mL/s} \cdot (60t\text{ s/min}) + 0.2\text{ L}. Determine the number of operational times tt (in minutes) at which both pumps deliver the exact same cumulative volume.

Answer

No solution. Convert to mL\text{mL}: the first pump delivers 40t+20040t + 200, and the second pump delivers (0.5×60)t+(0.2×1000)=30t+200(0.5 \times 60)t + (0.2 \times 1000) = 30t + 200. Equating them yields 40t+200=30t+200  ⟹  10t=0  ⟹  t=0 min40t + 200 = 30t + 200 \implies 10t = 0 \implies t = 0\text{ min} as a single unique operational time.

02

Two sensors track gas volume over tt minutes: Sensor A records VA=120t cm3+3.0 dm3V_A = 120t\text{ cm}^3 + 3.0\text{ dm}^3. Sensor B records VB=2.0×103 mm3/s⋅(60t)+3000 cm3V_B = 2.0\times 10^3\text{ mm}^3/\text{s}\cdot(60t) + 3000\text{ cm}^3. Determine whether VA=VBV_A = V_B has unique, no, or infinitely many solutions (1 dm3=1000 cm31\text{ dm}^3 = 1000\text{ cm}^3, 1 cm3=1000 mm31\text{ cm}^3 = 1000\text{ mm}^3).

Answer

Infinitely many solutions for all t≥0t \ge 0. In cm3\text{cm}^3: VA=120t+3000V_A = 120t + 3000. For VBV_B: (2000×60/1000)t+3000=120t+3000(2000 \times 60 / 1000)t + 3000 = 120t + 3000. Since VA=VBV_A = V_B is an identity, it holds for all non-negative tt.

03

An automated chemical reactor compares cumulative mass inputs: Stream 1 adds 0.18t kg+450 g0.18t\text{ kg} + 450\text{ g} over tt minutes. Stream 2 adds 50 mg/s⋅(60t s/min)+0.45 kg50\text{ mg/s}\cdot(60t\text{ s/min}) + 0.45\text{ kg}. Determine if there is a point in time t≥0t \ge 0 where Stream 1 equals Stream 2.

Answer

A unique solution at t=0t = 0. In grams: Stream 1 is 180t+450180t + 450. Stream 2 is (50×60/1000)t+450=3t+450(50 \times 60 / 1000)t + 450 = 3t + 450. Setting 180t+450=3t+450180t + 450 = 3t + 450 gives 177t=0177t = 0, so t=0 mint = 0\text{ min} is the unique non-negative solution.

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