1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 x 10

Q: Why do we use $R = 0.083$ instead of $8.314$?

The value of the ideal gas constant $R$ depends on the units of pressure and volume. Since pressure is given in bar and volume in liters, we must use $R = 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1}$. If pressure were in Pascals and volume in cubic meters, we would use $8.314$.

Q: Why did we convert $100 \text{ mL}$ to $0.1 \text{ L}$?

Molarity ($C$) is strictly defined as the number of moles of solute per **liter** of solution. Using milliliters directly in the formula $\pi = CRT$ will result in an answer that is off by a factor of $1000$.

Q: Why is osmotic pressure preferred for finding the molar mass of polymers?

Polymers have very large molar masses, which means even a decent mass of polymer results in a very small number of moles. This causes negligible changes in boiling or freezing points. However, osmotic pressure is sensitive enough to give a measurable reading even for such dilute solutions.

Q: What about the van't Hoff factor ($i$) for the biopolymer?

Biopolymers are generally large, stable covalent molecules that do not dissociate into ions in water. Therefore, they behave as non-electrolytes, and their van't Hoff factor $i$ is exactly $1$.

Q: How do I handle the calculation without making a mistake with the powers of $10$?

Group the numbers and the powers of $10$ separately. Calculate $\frac{1.46 \times 0.083 \times 300}{2.42 \times 0.1}$ first, which gives roughly $150.2$. Then handle the $10^{-3}$ in the denominator, which moves to the numerator as $10^3$, giving $150.2 \times 10^3$ or $15.02 \times 10^4$.

Question

1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of $2.42 	imes 10^{-3}$ bar. The molar mass of the biopolymer is ......... $	imes 10^4 	ext{ g mol}^{-1}$. (Round off to the nearest integer)
[Use : $R = 0.083 	ext{ L bar mol}^{-1} 	ext{ K}^{-1}$]

Sigma Percentile · Accepted Answer

### The Setup: A Macromolecular Mystery

Imagine a beaker filled with exactly $100 	ext{ mL}$ of water at a warm $300 	ext{ K}$. Into this, we dissolve $1.46 	ext{ g}$ of a biopolymer.

This creates a solution that exerts a tiny, yet measurable osmotic pressure of $2.42 	imes 10^{-3} 	ext{ bar}$. Our mission is to uncover the molar mass of this mysterious biopolymer.

### The Master Equation

To find the molar mass of this biopolymer, we need the master equation for osmotic pressure.

According to **van't Hoff's law** for dilute solutions, osmotic pressure $\pi$ is equal to the molar concentration $C$, times the universal gas constant $R$, times the absolute temperature $T$.

$$ \pi = CRT $$

### Expanding the Concentration

Now, what exactly is the concentration $C$? It is simply the number of moles of the solute $n$ divided by the volume of the solution in liters $V$.

And the number of moles is the given mass $w$ divided by the unknown molar mass $M$. Let us expand our formula to include these terms.

$$ C = \frac{n}{V} = \frac{w}{M \cdot V} $$

Substituting this back into our master equation, we get:

$$ \pi = \frac{w \cdot R \cdot T}{M \cdot V} $$

### Rearranging for the Unknown

Since we need to find the molar mass $M$, let us rearrange this equation. We will swap $M$ and $\pi$.

So, $M$ equals the mass $w$, times $R$, times $T$, all divided by the osmotic pressure $\pi$, times the volume $V$.

$$ M = \frac{w \cdot R \cdot T}{\pi \cdot V} $$

### The Crucial Substitution

Let us carefully plug in our known values. The mass is $1.46 	ext{ g}$. The gas constant $R$ is $0.083 	ext{ L bar K}^{-1} 	ext{ mol}^{-1}$. The temperature is $300 	ext{ K}$.

The osmotic pressure is $2.42 	imes 10^{-3} 	ext{ bar}$, and the volume must be converted to liters, which is $0.1 	ext{ L}$. Do not rush through this substitution; units are the most common trap here!

$$ M = \frac{1.46 	imes 0.083 	imes 300}{2.42 	imes 10^{-3} 	imes 0.1} $$

### The Final Calculation

Now for the calculation. The numerator, $1.46 	imes 0.083 	imes 300$, gives us $36.354$.

The denominator, $2.42 	imes 10^{-3} 	imes 0.1$, is $2.42 	imes 10^{-4}$.

Dividing these gives us approximately $150223 	ext{ g mol}^{-1}$.

$$ M = \frac{36.354}{2.42 	imes 10^{-4}} = 150223 	ext{ g mol}^{-1} $$

### Formatting the Answer

The question asks for the answer in the format of something $	imes 10^4$. So, we write $150223$ as $15.02 	imes 10^4$.

Rounding off to the nearest integer, we get $15$. Our final answer is **$15$**.

### Why Osmotic Pressure?

Think about this: why do we use osmotic pressure to find the molar mass of biopolymers instead of other colligative properties like freezing point depression?

It is because biopolymers have huge molar masses, making changes in freezing point too tiny to measure accurately. Osmotic pressure, on the other hand, yields a significantly measurable value even for very dilute solutions!

Enter Numerical Value:

Visualized Solution

$w = 1.46 g, V = 0.1 L$

$π = C R T$

$C = \frac{w}{M \cdot V}$

$M = \frac{w \cdot R \cdot T}{π \cdot V}$

$M = \frac{1.46 \times 0.083 \times 300}{2.42 \times 1 0 ^{- 3} \times 0.1}$

$M = 150223 g mol^{- 1}$

$M = 15.02 \times 1 0^{4} g mol^{- 1}$

$Why Osmotic Pressure?$

The Sigma Insight: Colligative Properties

The Setup

The Master Equation

Expanding the Concentration

Rearranging for the Unknown

The Crucial Substitution

The Final Calculation

Formatting the Answer

Why Osmotic Pressure?

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