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Osmotic Pressure of Solutions of Electrolytes

Environmental Engineering · FE Reference Handbook section

Environmental Engineering
8 formulas
10 exam-style examples
~60 min
All Environmental Engineering lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Osmotic Pressure of Solutions of Electrolytes — solve for osmotic pressure — Osmotic Pressure of Solutions of Electrolytes

osmotic pressure of solutions of electrolytes for a saline feedwater Given number of ions per molecule (Phi) = 3.0000; molar concentration (M) = 0.3800 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 281.0 K, determine the osmotic pressure (pi) in atm.

Given

  • numberofionspermolecule(Phi)=3.0000number of ions per molecule (Phi) = 3.0000
  • molarconcentration(M)=0.3800mol/Lmolar concentration (M) = 0.3800 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=281.0Kabsolute temperature (T) = 281.0 K

Find

osmotic pressure (pi), in atm

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except pi is given, so isolate pi symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for pi:

    π=ΦMRT\pi = \Phi M R T
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 3.0000, molar concentration (M) = 0.3800 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 281.0 K.

  4. Step 4 — Substitute the given values:

    π=3.00000.38000.0821281.0\pi = 3.0000 0.3800 0.0821 281.0
  5. Step 5 — Evaluate:

    π=26.2999 atm\pi = 26.2999\ \text{atm}
  6. Step 6 — Check: returning pi = 26.2999 atm to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
π=26.2999 atm\pi = 26.2999\ \text{atm}

Why the other options are there

  • 52.5998 — kept a factor of two that cancels in the correct rearrangement.
  • 13.1500 — dropped that same factor in the other direction.
  • 28.9299 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 2
Osmotic Pressure of Solutions of Electrolytes — solve for molar concentration — Osmotic Pressure of Solutions of Electrolytes (2)

osmotic pressure calculation for an electrolyte solution in RO design Given number of ions per molecule (Phi) = 2.0000; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 300.0 K; osmotic pressure (pi) = 57.4800 atm, determine the molar concentration (M) in mol/L.

Given

  • numberofionspermolecule(Phi)=2.0000number of ions per molecule (Phi) = 2.0000
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=300.0Kabsolute temperature (T) = 300.0 K
  • osmoticpressure(pi)=57.4800atmosmotic pressure (pi) = 57.4800 atm

Find

molar concentration (M), in mol/L

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for M:

    M=πΦRTM = \dfrac{\pi}{\Phi R T}
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 2.0000, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 300.0 K, osmotic pressure (pi) = 57.4800 atm.

  4. Step 4 — Substitute the given values:

    M=57.48002.00000.0821300.0M = \dfrac{57.4800}{2.0000 0.0821 300.0}
  5. Step 5 — Evaluate:

    M=1.1669 mol/LM = 1.1669\ \text{mol/L}
  6. Step 6 — Check: returning M = 1.1669 mol/L to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
M=1.1669 mol/LM = 1.1669\ \text{mol/L}

Why the other options are there

  • 2.3337 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5834 — dropped that same factor in the other direction.
  • 1.2836 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 3
Osmotic Pressure of Solutions of Electrolytes — solve for absolute temperature — Osmotic Pressure of Solutions of Electrolytes (3)

osmotic pressure of an electrolyte feed to a membrane system Given number of ions per molecule (Phi) = 3.0000; molar concentration (M) = 0.4310 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; osmotic pressure (pi) = 41.0200 atm, determine the absolute temperature (T) in K.

Given

  • numberofionspermolecule(Phi)=3.0000number of ions per molecule (Phi) = 3.0000
  • molarconcentration(M)=0.4310mol/Lmolar concentration (M) = 0.4310 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • osmoticpressure(pi)=41.0200atmosmotic pressure (pi) = 41.0200 atm

Find

absolute temperature (T), in K

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for T:

    T=πΦMRT = \dfrac{\pi}{\Phi M R}
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 3.0000, molar concentration (M) = 0.4310 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, osmotic pressure (pi) = 41.0200 atm.

  4. Step 4 — Substitute the given values:

    T=41.02003.00000.43100.0821T = \dfrac{41.0200}{3.0000 0.4310 0.0821}
  5. Step 5 — Evaluate:

    T=386.4 KT = 386.4\ \text{K}
  6. Step 6 — Check: returning T = 386.4 K to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
T=386.4 KT = 386.4\ \text{K}

Why the other options are there

  • 772.8 — kept a factor of two that cancels in the correct rearrangement.
  • 193.2 — dropped that same factor in the other direction.
  • 425.1 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 4
Osmotic Pressure of Solutions of Electrolytes — solve for osmotic pressure (case 2) — Osmotic Pressure of Solutions of Electrolytes (4)

osmotic pressure of solutions of electrolytes for a saline feedwater Given number of ions per molecule (Phi) = 4.0000; molar concentration (M) = 0.0950 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 286.0 K, determine the osmotic pressure (pi) in atm.

Given

  • numberofionspermolecule(Phi)=4.0000number of ions per molecule (Phi) = 4.0000
  • molarconcentration(M)=0.0950mol/Lmolar concentration (M) = 0.0950 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=286.0Kabsolute temperature (T) = 286.0 K

Find

osmotic pressure (pi), in atm

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except pi is given, so isolate pi symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for pi:

    π=ΦMRT\pi = \Phi M R T
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 4.0000, molar concentration (M) = 0.0950 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 286.0 K.

  4. Step 4 — Substitute the given values:

    π=4.00000.09500.0821286.0\pi = 4.0000 0.0950 0.0821 286.0
  5. Step 5 — Evaluate:

    π=8.9226 atm\pi = 8.9226\ \text{atm}
  6. Step 6 — Check: returning pi = 8.9226 atm to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
π=8.9226 atm\pi = 8.9226\ \text{atm}

Why the other options are there

  • 17.8453 — kept a factor of two that cancels in the correct rearrangement.
  • 4.4613 — dropped that same factor in the other direction.
  • 9.8149 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 5
Osmotic Pressure of Solutions of Electrolytes — solve for molar concentration (case 2) — Osmotic Pressure of Solutions of Electrolytes (5)

osmotic pressure calculation for an electrolyte solution in RO design Given number of ions per molecule (Phi) = 3.0000; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 278.0 K; osmotic pressure (pi) = 25.2200 atm, determine the molar concentration (M) in mol/L.

Given

  • numberofionspermolecule(Phi)=3.0000number of ions per molecule (Phi) = 3.0000
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=278.0Kabsolute temperature (T) = 278.0 K
  • osmoticpressure(pi)=25.2200atmosmotic pressure (pi) = 25.2200 atm

Find

molar concentration (M), in mol/L

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for M:

    M=πΦRTM = \dfrac{\pi}{\Phi R T}
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 3.0000, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 278.0 K, osmotic pressure (pi) = 25.2200 atm.

  4. Step 4 — Substitute the given values:

    M=25.22003.00000.0821278.0M = \dfrac{25.2200}{3.0000 0.0821 278.0}
  5. Step 5 — Evaluate:

    M=0.3683 mol/LM = 0.3683\ \text{mol/L}
  6. Step 6 — Check: returning M = 0.3683 mol/L to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
M=0.3683 mol/LM = 0.3683\ \text{mol/L}

Why the other options are there

  • 0.7367 — kept a factor of two that cancels in the correct rearrangement.
  • 0.1842 — dropped that same factor in the other direction.
  • 0.4052 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 6
Osmotic Pressure of Solutions of Electrolytes — solve for absolute temperature (case 2) — Osmotic Pressure of Solutions of Electrolytes (6)

osmotic pressure of an electrolyte feed to a membrane system Given number of ions per molecule (Phi) = 1.0000; molar concentration (M) = 0.3890 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; osmotic pressure (pi) = 30.0300 atm, determine the absolute temperature (T) in K.

Given

  • numberofionspermolecule(Phi)=1.0000number of ions per molecule (Phi) = 1.0000
  • molarconcentration(M)=0.3890mol/Lmolar concentration (M) = 0.3890 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • osmoticpressure(pi)=30.0300atmosmotic pressure (pi) = 30.0300 atm

Find

absolute temperature (T), in K

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for T:

    T=πΦMRT = \dfrac{\pi}{\Phi M R}
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 1.0000, molar concentration (M) = 0.3890 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, osmotic pressure (pi) = 30.0300 atm.

  4. Step 4 — Substitute the given values:

    T=30.03001.00000.38900.0821T = \dfrac{30.0300}{1.0000 0.3890 0.0821}
  5. Step 5 — Evaluate:

    T=940.3 KT = 940.3\ \text{K}
  6. Step 6 — Check: returning T = 940.3 K to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
T=940.3 KT = 940.3\ \text{K}

Why the other options are there

  • 1,881 — kept a factor of two that cancels in the correct rearrangement.
  • 470.1 — dropped that same factor in the other direction.
  • 1,034 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 7
Osmotic Pressure of Solutions of Electrolytes — solve for osmotic pressure (case 3) — Osmotic Pressure of Solutions of Electrolytes (7)

osmotic pressure of solutions of electrolytes for a saline feedwater Given number of ions per molecule (Phi) = 2.0000; molar concentration (M) = 0.1530 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 274.0 K, determine the osmotic pressure (pi) in atm.

Given

  • numberofionspermolecule(Phi)=2.0000number of ions per molecule (Phi) = 2.0000
  • molarconcentration(M)=0.1530mol/Lmolar concentration (M) = 0.1530 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=274.0Kabsolute temperature (T) = 274.0 K

Find

osmotic pressure (pi), in atm

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except pi is given, so isolate pi symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for pi:

    π=ΦMRT\pi = \Phi M R T
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 2.0000, molar concentration (M) = 0.1530 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 274.0 K.

  4. Step 4 — Substitute the given values:

    π=2.00000.15300.0821274.0\pi = 2.0000 0.1530 0.0821 274.0
  5. Step 5 — Evaluate:

    π=6.8836 atm\pi = 6.8836\ \text{atm}
  6. Step 6 — Check: returning pi = 6.8836 atm to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
π=6.8836 atm\pi = 6.8836\ \text{atm}

Why the other options are there

  • 13.7672 — kept a factor of two that cancels in the correct rearrangement.
  • 3.4418 — dropped that same factor in the other direction.
  • 7.5720 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 8
Osmotic Pressure of Solutions of Electrolytes — solve for molar concentration (case 3) — Osmotic Pressure of Solutions of Electrolytes (8)

osmotic pressure calculation for an electrolyte solution in RO design Given number of ions per molecule (Phi) = 2.0000; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 301.0 K; osmotic pressure (pi) = 51.6300 atm, determine the molar concentration (M) in mol/L.

Given

  • numberofionspermolecule(Phi)=2.0000number of ions per molecule (Phi) = 2.0000
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=301.0Kabsolute temperature (T) = 301.0 K
  • osmoticpressure(pi)=51.6300atmosmotic pressure (pi) = 51.6300 atm

Find

molar concentration (M), in mol/L

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for M:

    M=πΦRTM = \dfrac{\pi}{\Phi R T}
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 2.0000, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 301.0 K, osmotic pressure (pi) = 51.6300 atm.

  4. Step 4 — Substitute the given values:

    M=51.63002.00000.0821301.0M = \dfrac{51.6300}{2.0000 0.0821 301.0}
  5. Step 5 — Evaluate:

    M=1.0446 mol/LM = 1.0446\ \text{mol/L}
  6. Step 6 — Check: returning M = 1.0446 mol/L to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
M=1.0446 mol/LM = 1.0446\ \text{mol/L}

Why the other options are there

  • 2.0893 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5223 — dropped that same factor in the other direction.
  • 1.1491 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 9
Osmotic Pressure of Solutions of Electrolytes — solve for absolute temperature (case 3) — Osmotic Pressure of Solutions of Electrolytes (9)

osmotic pressure of an electrolyte feed to a membrane system Given number of ions per molecule (Phi) = 1.0000; molar concentration (M) = 0.3350 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; osmotic pressure (pi) = 0.9400 atm, determine the absolute temperature (T) in K.

Given

  • numberofionspermolecule(Phi)=1.0000number of ions per molecule (Phi) = 1.0000
  • molarconcentration(M)=0.3350mol/Lmolar concentration (M) = 0.3350 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • osmoticpressure(pi)=0.9400atmosmotic pressure (pi) = 0.9400 atm

Find

absolute temperature (T), in K

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for T:

    T=πΦMRT = \dfrac{\pi}{\Phi M R}
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 1.0000, molar concentration (M) = 0.3350 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, osmotic pressure (pi) = 0.9400 atm.

  4. Step 4 — Substitute the given values:

    T=0.94001.00000.33500.0821T = \dfrac{0.9400}{1.0000 0.3350 0.0821}
  5. Step 5 — Evaluate:

    T=34.1775 KT = 34.1775\ \text{K}
  6. Step 6 — Check: returning T = 34.1775 K to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
T=34.1775 KT = 34.1775\ \text{K}

Why the other options are there

  • 68.3549 — kept a factor of two that cancels in the correct rearrangement.
  • 17.0887 — dropped that same factor in the other direction.
  • 37.5952 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

Example 10
Osmotic Pressure of Solutions of Electrolytes — solve for osmotic pressure (case 4) — Osmotic Pressure of Solutions of Electrolytes (10)

osmotic pressure of solutions of electrolytes for a saline feedwater Given number of ions per molecule (Phi) = 4.0000; molar concentration (M) = 0.0420 mol/L; gas constant (R) = 0.0821 L*atm/mol/K; absolute temperature (T) = 304.0 K, determine the osmotic pressure (pi) in atm.

Given

  • numberofionspermolecule(Phi)=4.0000number of ions per molecule (Phi) = 4.0000
  • molarconcentration(M)=0.0420mol/Lmolar concentration (M) = 0.0420 mol/L
  • gasconstant(R)=0.0821L∗atm/mol/Kgas constant (R) = 0.0821 L*atm/mol/K
  • absolutetemperature(T)=304.0Kabsolute temperature (T) = 304.0 K

Find

osmotic pressure (pi), in atm

Start with the thinking

  • The governing relation printed in this handbook section is Osmotic Pressure of Solutions of Electrolytes.
  • Everything except pi is given, so isolate pi symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Osmotic pressure of solutions of electrolytes determines the driving pressure that must be overcome in reverse osmosis membranes.

Step-by-step solution

  1. Step 1 — State the governing relation:

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T
  2. Step 2 — Rearrange symbolically for pi:

    π=ΦMRT\pi = \Phi M R T
  3. Step 3 — List the givens: number of ions per molecule (Phi) = 4.0000, molar concentration (M) = 0.0420 mol/L, gas constant (R) = 0.0821 L*atm/mol/K, absolute temperature (T) = 304.0 K.

  4. Step 4 — Substitute the given values:

    π=4.00000.04200.0821304.0\pi = 4.0000 0.0420 0.0821 304.0
  5. Step 5 — Evaluate:

    π=4.1930 atm\pi = 4.1930\ \text{atm}
  6. Step 6 — Check: returning pi = 4.1930 atm to

    π=Φ×M×R×T\pi = \Phi \times M \times R \times T

    reproduces the given quantities, and both sides carry the same units.

Answer:
π=4.1930 atm\pi = 4.1930\ \text{atm}

Why the other options are there

  • 8.3860 — kept a factor of two that cancels in the correct rearrangement.
  • 2.0965 — dropped that same factor in the other direction.
  • 4.6123 — rounded an intermediate value before the final step.

Reference: FE Handbook — Osmotic Pressure of Electrolyte Solutions

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