Q.Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37$^\circ$C.

The key idea is that osmotic pressure ($\Pi$) for a dilute solution follows van't Hoff's law: $\Pi = iCRT$, and for a non-electrolyte polymer, $i = 1$.

Step 1: Find the molar concentration $C$.

Mass of polymer = 1.0 g, molar mass $M = 185,000$ g/mol.

Moles of polymer = $\frac{1.0}{185,000} = 5.405 \times 10^{-6}$ mol.

Volume of solution = 450 mL = $0.450$ L.

$C = \frac{5.405 \times 10^{-6}}{0.450} = 1.201 \times 10^{-5}$ mol/L.

Step 2: Convert to SI units.

$C$ in mol/m$^3$: $1.201 \times 10^{-5}$ mol/L = $1.201 \times 10^{-2}$ mol/m$^3$ (since 1 L = $10^{-3}$ m$^3$).

Temperature $T = 37^\circ$C = $310$ K. …

Why osmotic pressure works for molar mass

Osmotic pressure is a colligative property - it depends solely on the concentration of solute particles, not on their chemical nature. For a non-electrolyte like a polymer, each molecule contributes one particle, so $i = 1$. This makes osmotic pressure ideal for finding the molar mass of large molecules: even a tiny mass of polymer gives a measurable pressure, whereas boiling point elevation or freezing point depression would be too small to detect.

The governing equation is:

where $\Pi$ is osmotic pressure (Pa), $w$ is mass of solute (g), $M$ is molar mass (g/mol), $V$ is volume of solution (m$^3$), $R$ is the gas constant, and $T$ is absolute temperature (K).

Step-by-step calculation

1. Convert temperature to Kelvin

2. Use SI gas constant

3. Convert volume to cubic metres

$V = 450 \times 10^{-6} = 4.50 \times 10^{-4}\ \text{m}^3$ …

Method: Osmotic Pressure Formula (van't Hoff Equation)

This method uses the direct relationship between osmotic pressure and molar concentration for non-electrolyte solutions.

Steps

1. Identify the formula The van't Hoff equation for osmotic pressure is:

For a non-electrolyte polymer, $i = 1$, so:

2. Convert temperature to Kelvin

3. Calculate the molar concentration ($C$)
   • Moles of polymer:

• Volume in litres:

• Molarity:

4. Use the value of $R$ in SI units For pressure in pascals (Pa), use:

Note: $1\ \text{L} = 10^{-3}\ \text{m}^3$, so $C$ in $\text{mol/m}^3$ is: …

Let’s first solve it correctly, then list the common mistakes and how to avoid each.

✓ Correct Solution (for reference)

Given:

• Mass of polymer, $w = 1.0 \text{ g}$
• Molar mass, $M = 185{,}000 \text{ g mol}^{-1}$
• Volume of solution, $V = 450 \text{ mL} = 0.450 \text{ L}$
• Temperature, $T = 37^\circ \text{C} = 37 + 273 = 310 \text{ K}$
• Gas constant, $R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}$ (for atm) or $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$ (for Pa)

Step 1: Number of moles

Step 2: Molarity

Step 3: Osmotic pressure (in Pa)

Use $\Pi = CRT$ with $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$ and $C$ in $\text{mol m}^{-3}$.

Convert molarity to $\text{mol m}^{-3}$:

So,

Now,

Final answer: $\boxed{30.95 \text{ Pa}}$ (approximately 31 Pa)

✗ Common Mistakes & How to Avoid Them

1. Forgetting to convert temperature to Kelvin

• Mistake: Using $T = 37^\circ\text{C}$ directly in $\Pi = CRT$.
• Why it’s wrong: The gas constant $R$ has units per Kelvin — using Celsius gives a completely wrong (and meaningless) result.
• How to avoid: Always add 273 to Celsius: $T(\text{K}) = T(^\circ\text{C}) + 273$. For 37°C, it’s 310 K.

2. Using wrong units for volume

• Mistake: Plugging $V = 450 \text{ mL}$ directly into $C = n/V$ without converting to litres.
• Why it’s wrong: Molarity is moles per litre, not per mL.
• How to avoid: Convert mL to L by dividing by 1000: $450 \text{ mL} = 0.450 \text{ L}$.

3. Confusing molar mass with molecular mass in g/mol

• Mistake: Treating 185,000 as the mass of one molecule (in amu) and then using it incorrectly.
• Why it’s wrong: Molar mass is already in g/mol — no further conversion needed.
• How to avoid: Remember: molar mass (g/mol) = molecular mass (amu) numerically. Just use it directly in $n = w/M$.

4. Using the wrong value of $R$ for the required unit of pressure

• Mistake: Using $R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}$ and then reporting pressure in pascals without converting.
• Why it’s wrong: That $R$ gives pressure in atm, not Pa.
• How to avoid:
  • If answer needed in Pa, use $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$ and ensure concentration is in $\text{mol m}^{-3}$.
  • If you use $R = 0.0821$, convert atm to Pa: $1 \text{ atm} = 101325 \text{ Pa}$.

5. Forgetting to convert molarity to $\text{mol m}^{-3}$ when using $R = 8.314$

• Mistake: Plugging $C$ in $\text{mol L}^{-1}$ directly into $\Pi = CRT$ with $R = 8.314$. …