An aqeous solution containing 1g/L of a polymer exerts osmotic pressure of 4 torr at 300k R = 0.0821the molar mass of the polymer is?

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I had solve this okk friends

Answer

Calculating the Molar Mass of the Polymer
- Given:
- Osmotic pressure (π) = 4 torr
- Concentration of the polymer (c) = 1 g/L
- R = 0.0821 L.atm/mol.K
- Temperature (T) = 300 K
- We can use the formula for osmotic pressure:
π = MRT
where:
π = osmotic pressure
M = molarity of the solution
R = gas constant
T = temperature in Kelvin
- Since the concentration is given in grams per liter, we need to convert it to molarity by dividing by the molar mass.
- Rearranging the formula to solve for M (molarity):
M = π / RT
- Substituting the given values:
M = 4 torr / (0.0821 L.atm/mol.K * 300 K)
- Calculate the molarity of the solution.
- Once we have the molarity, we can determine the molar mass of the polymer by using the formula:
Molar mass = Mass / Number of moles
- Since we know the concentration and molar mass are related by:
Molar mass = Mass / (Molarity * Volume)
- Substituting the values, we can calculate the molar mass of the polymer.
- Therefore, by following these steps and calculations, we can determine the molar mass of the polymer in the given aqueous solution.

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