The main application of osmotic pressure measurement is in the determination of the molar mass of a substance which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and colloids.This is due to the fact that even a very small concentraion of the solution produces fairly large magnitude of osomotic pressure.In the laboratory the concentrations usually employed are of the order of `10^(-3)` to `10^(-4)` M.At concentration of `10^(-3)` mol `dm^(-3)` , the magnitude of osmotic pressure of 300 K is :
`P=10^(-3)xx0.082xx300=0.0246` atm
or `0.0246xx1.01325xx10^5=2492.595 Pa`
At this concentration, the values of other colligative properties such as boiling point elevation and depression in freezing point are too small to be determined experimentally.
Further polymers have following two types of molar masses :
(A) Number average molar mass `(barM_n)`, which is given by `(undersetisumN_iM_i)/(undersetisumN_i)`
where `N_i` is the number of molecules having molar mass `M_i`.
(B) Molar average molar mass `(barM_m)`, which is given by `(undersetisumN_iM_i^2)/(undersetisumN_iM_i)`
Obviously the former is independent of the individual characteristics of the molecules and gives equal weightage to large and small molecules in the polymer sample.On the other hand later gives more weightage to the heavier molecules.Infact with the help of a colligative property only one type of molar mass of the polymer can be determined.
One gram each of polymer A (molar mass=2000) and B(molar mass=6000) is dissolved in water to form one litre solution at `27^@C`.The osmotic pressure of this solution will be :

Options

A. 0.0164 atm
B. 1862 Pa
C. Both are correct
D. None of these

Explanation

To solve the problem of determining the osmotic pressure of a solution containing one gram each of two polymers (A and B) with molar masses of 2000 g/mol and 6000 g/mol, respectively, we can follow these steps:

### Step 1: Calculate the number of moles of each polymer.

1. **For Polymer A:**
- Molar mass of A = 2000 g/mol
- Mass of A = 1 g
- Number of moles of A (n_A) = Mass / Molar mass = \( \frac{1 \text{ g}}{2000 \text{ g/mol}} = 0.0005 \text{ mol} \)

2. **For Polymer B:**
- Molar mass of B = 6000 g/mol
- Mass of B = 1 g
- Number of moles of B (n_B) = Mass / Molar mass = \( \frac{1 \text{ g}}{6000 \text{ g/mol}} = 0.0001667 \text{ mol} \)

### Step 2: Calculate the total number of moles in the solution.

- Total number of moles (n_total) = n_A + n_B
- \( n_{\text{total}} = 0.0005 \text{ mol} + 0.0001667 \text{ mol} = 0.0006667 \text{ mol} \)

### Step 3: Calculate the concentration of the solution.

- Since the total volume of the solution is 1 liter, the concentration (C) can be calculated as:
- \( C = \frac{n_{\text{total}}}{\text{Volume}} = \frac{0.0006667 \text{ mol}}{1 \text{ L}} = 0.0006667 \text{ M} \)

### Step 4: Use the osmotic pressure formula.

The formula for osmotic pressure (P) is given by:
\[ P = C \cdot R \cdot T \]
Where:
- \( C \) = concentration in mol/L
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin = 27°C + 273 = 300 K

### Step 5: Substitute the values into the formula.

- Now substituting the values:
\[ P = 0.0006667 \text{ M} \cdot 0.0821 \text{ L·atm/(K·mol)} \cdot 300 \text{ K} \]

### Step 6: Calculate the osmotic pressure.

- Performing the calculation:
\[ P = 0.0006667 \cdot 0.0821 \cdot 300 \]
\[ P = 0.0164 \text{ atm} \]

### Step 7: Convert atm to Pa (if required).

- \( 1 \text{ atm} = 101325 \text{ Pa} \)
- Thus, \( P = 0.0164 \text{ atm} \cdot 101325 \text{ Pa/atm} \approx 1665.57 \text{ Pa} \)

### Final Answer:
The osmotic pressure of the solution is approximately **0.0164 atm** or **1665.57 Pa**.

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