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What is the osmotic pressure of a \(0.135 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) at \(20^{\circ} \mathrm{C} ?\) (Assume complete dissociation.)

Short Answer

Expert verified
Answer: The osmotic pressure of the 0.135 M Na2SO4 solution at 20°C is 9.86 atm.

Step by step solution

01

Identify the formula for osmotic pressure

We will use the formula for osmotic pressure (Π) given by the following equation: Π = nCRT, where n represents the number of moles of solute dissolved in a liter of solvent, C is the concentration of the solution in molarity, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
02

Convert the temperature to Kelvin

To calculate the osmotic pressure, we need to convert the given temperature, 20°C, into Kelvin. To do this, add 273.15 to the given temperature: T(K) = T(°C) + 273.15 = 20°C + 273.15 = 293.15 K.
03

Determine the number of ions in the solution after dissociation

Since Na2SO4 dissociates completely in the solution, it will separate into ions: Na2SO4 → 2Na+ + SO4²-. Thus, for each formula unit of Na2SO4, there will be 3 particles. Hence, the effective concentration of particles in the solution will be 3 times the given concentration of 0.135 M.
04

Calculate the effective concentration of particles

Multiply the given concentration by the number of particles: Effective concentration = 0.135 M × 3 = 0.405 M.
05

Calculate the osmotic pressure using the formula

Now we will use the formula Π = nCRT with the effective concentration derived in step 4 and the temperature in Kelvin from step 2. Osmotic pressure, Π = (0.405 mol/L) × (0.0821 L atm/mol K) × (293.15 K) = 9.86 atm. So, the osmotic pressure of the 0.135 M Na2SO4 solution at 20°C is 9.86 atm.

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Most popular questions from this chapter

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