At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in t

Q: Why do we multiply the concentration by 1000?

The given concentration is in $\text{g/dm}^3$ (which is $\text{g/L}$). When we divide by molar mass, we get $\text{mol/L}$. However, standard SI unit for volume is $\text{m}^3$. Since $1 \text{ m}^3 = 1000 \text{ L}$, we multiply by 1000 to convert $\text{mol/L}$ to $\text{mol/m}^3$.

Q: Can I use $R = 0.0821 \text{ L atm / (K mol)}$ instead?

Yes, you can! But you would first need to convert the hydrostatic pressure from Pascals to atmospheres by dividing by $1.013 \times 10^5$. Using SI units throughout is generally safer and avoids messy conversion factors.

Q: Why is the density of the solution taken as $1.00 \text{ g/cm}^3$?

For very dilute aqueous solutions, the amount of solute is so small that the density of the solution is practically equal to the density of pure water, which is $1.00 \text{ g/cm}^3$.

Q: What is a macromolecule?

A macromolecule is a very large molecule, such as a protein, polymer, or nucleic acid, typically composed of thousands of atoms. They have very high molar masses, which is why their solutions are very dilute in terms of molarity.

Q: Why don't we use freezing point depression for macromolecules?

Because macromolecules have huge molar masses, their molar concentration in solution is extremely tiny. This results in a freezing point depression ($\Delta T_f$) that is too small to measure accurately. Osmotic pressure, however, yields a significantly large and measurable value even at such low concentrations.

Question

At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = $1.00 	ext{ g cm}^{-3}$) where h is equal to $2.00 	ext{ cm}$. If the concentration of the dilute solution of the macromolecule is $2.00 	ext{ g dm}^{-3}$, the molar mass of the macromolecule is calculated to be $X 	imes 10^4 	ext{ g mol}^{-1}$. The value of $X$ is ______.
Use : Universal gas constant (R) = $8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}$ and acceleration due to gravity (g) = $10 	ext{ m s}^{-2}$

Sigma Percentile · Accepted Answer

### The Setup: Visualizing Osmotic Pressure

Imagine an osmometer setup in a laboratory. You have a solution containing a large macromolecule separated from a pure solvent by a semi-permeable membrane. Because nature loves equilibrium, solvent molecules rush across the membrane to dilute the solution—a process we call osmosis.

As the solvent enters, the solution level rises in a vertical tube, creating a liquid column of height $h$. Gravity pulls down on this column, creating a hydrostatic pressure. When this downward hydrostatic pressure exactly balances the inward drive of osmosis, the system reaches equilibrium. This balancing pressure is precisely what we define as the **osmotic pressure** ($\pi$).

### The First Pillar: Hydrostatic Pressure

Our first mathematical tool is the classic fluid mechanics formula for hydrostatic pressure:

$$\pi = ho g h$$

Before we blindly plug in numbers, we must respect the units. This is where many students fall into a trap. To get our pressure in standard SI units (Pascals), every variable must be in SI units.

The density $ho$ is given as $1.00 	ext{ g/cm}^3$. Converting this to kilograms per cubic meter gives us $10^3 	ext{ kg/m}^3$. The height $h$ is $2.00 	ext{ cm}$, which translates to $2 	imes 10^{-2} 	ext{ m}$. The acceleration due to gravity $g$ is given as $10 	ext{ m/s}^2$.

Substituting these pristine SI values:

$$\pi = (10^3) 	imes (10) 	imes (2 	imes 10^{-2}) = 200 	ext{ Pa}$$

We now have a clean, measurable osmotic pressure of $200 	ext{ Pa}$.

### The Second Pillar: Van't Hoff Equation

How does this pressure connect to the elusive molar mass of our macromolecule? Enter the Van't Hoff equation for dilute solutions, which beautifully mirrors the ideal gas law:

$$\pi = C R T$$

Here, $C$ is the molar concentration, $R$ is the universal gas constant, and $T$ is the absolute temperature.

We are given the mass concentration as $2.00 	ext{ g/dm}^3$. Since $1 	ext{ dm}^3$ is exactly $1 	ext{ L}$, this is $2.00 	ext{ g/L}$. To find the molarity (moles per liter), we divide by the unknown molar mass $M$, giving us $\frac{2.00}{M} 	ext{ mol/L}$.

But wait! Our pressure is in Pascals (SI), and our gas constant $R$ is $8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}$ (SI). We cannot use liters! We must convert our concentration to moles per cubic meter. Since there are $1000$ liters in a cubic meter, we multiply by $1000$:

$$C = \frac{2.00}{M} 	imes 1000 = \frac{2000}{M} 	ext{ mol/m}^3$$

### The Grand Equivalence: Solving for Molar Mass

Now, we bring our two pillars together. We substitute our calculated $\pi$ and our expression for $C$ into the Van't Hoff equation:

$$200 = \left( \frac{2000}{M} ight) 	imes 8.3 	imes 300$$

Look at the elegance of this equation. We have isolated a single unknown, $M$. Let's rearrange to solve for it:

$$M = \frac{2000 	imes 8.3 	imes 300}{200}$$

The $200$ in the denominator cancels perfectly with the $2000$ in the numerator, leaving a factor of $10$:

$$M = 10 	imes 8.3 	imes 300$$

$$M = 83 	imes 300 = 24900 	ext{ g/mol}$$

### The Final Polish

The question demands the answer in a specific scientific notation format: $X 	imes 10^4 	ext{ g mol}^{-1}$.

We rewrite our result:

$$24900 = 2.49 	imes 10^4$$

Comparing this to the requested format, we find our final answer:

**$$X = 2.49$$**

As a parting thought, realize *why* we use osmotic pressure for macromolecules. Proteins and polymers have massive molar masses, meaning their solutions have incredibly tiny molar concentrations. Colligative properties like boiling point elevation would yield unmeasurably small changes. Osmotic pressure, however, amplifies this tiny concentration into a macroscopic, easily measurable fluid height at room temperature. It is a triumph of physical chemistry!

Enter Numerical Value:

Visualized Solution

$Osmotic Pressure Setup$

$π = ρ g h$

$Unit Conversions$

$Calculating π$

$π = C R T$

$Molar Concentration (C)$

$Substituting into Van’t Hoff$

$Solving for M$

$Final Answer$

$Why Osmotic Pressure?$

The Sigma Insight: Colligative Properties

The Setup

The First Pillar

The Second Pillar

The Grand Equivalence

The Final Polish

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Osmotic Pressure Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } π=ρgh

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Unit Conversions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating π

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } π=CRT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molar Concentration (C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting into Van’t Hoff

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Why Osmotic Pressure?