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For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic pressure at 300 K.[R = 0·0821 L atm K–1 mol–1

Question

For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic pressure at 300 K.

R=0.0821 L atm K1mol1 R = 0.0821 \ \text{L atm K}^{-1} \text{mol}^{-1}

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Solution

Sure, here's how you can calculate the osmotic pressure for a 5% solution of urea at 300 K:

Step 1: Calculate the molarity of the solution. The 5% solution means that there are 5 g of urea in 100 g of solution. Since the density of water is approximately 1 g/mL, we can assume that 100 g of solution is approximately 100 mL or 0.1 L.

So, the molarity (M) = mass of solute (g) / molar mass (g/mol) / volume of solution (L) M = 5 g / 60 g/mol / 0.1 L = 0.833 mol/L

Step 2: Use the formula for osmotic pressure. The formula for osmotic pressure (π) is π = MRT, where M is the molarity, R is the gas constant, and T is the temperature in Kelvin.

So, π = 0.833 mol/L * 0.0821 L atm K–1 mol–1 * 300 K = 20.45 atm

Therefore, the osmotic pressure of a 5% solution of urea at 300 K is 20.45 atm.

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