What is the osmotic pressure of a solution containing a solute with a molecular weight of75,000 at a concentration of 30 g L-1 and a solute with a molecular weight of 20,000 at a
concentration of 10 g L-1? The solute with a molecular weight of 75,000 is unstable and completely dissociates into four solutes of equal size.

Answers

Answer 1

The osmotic pressure of a solution containing a solute with a molecular weight of 75,000 at a concentration of 30 g L-1 and a solute with a molecular weight of 20,000 at a concentration of 10 g L-1 and the solute with a molecular weight of 75,000 is unstable and completely dissociates into four solutes of equal size is calculated using the following .

steps;Step 1: Calculate the number of solute particles present in 30 g L-1 of the 75,000 g mol-1 solute that dissociates into four equal solutesIf the 75,000 g mol-1 solute dissociates completely into four equal solutes, each with a molecular weight of 75,000/4=18,750 g mol-1; then the number of solute particles in 30 g L-1 of the 75,000 g mol-1 solute will be:Number of solute particles = (30/75,000) × 6.022 × 1023 × 4= 0.00009622 × 1024 = 9.622 × 1018 particlesStep 2: Calculate the number of solute particles present in 10 g L-1 of the 20,000 g mol-1 soluteNumber of solute particles = (10/20,000) × 6.022 × 1023= 0.0003011 × 1024 = 3.011 × 1019 particlesStep 3: Calculate the total number of solute particles in the solutionTotal number of solute particles = 9.622 × 1018 + 3.011 × 1019= 3.9732 × 1019 particlesStep 4: Calculate the molar concentration of the solutionMolar mass of 75,000 g mol-1 solute (dissociated) = 18,750 g mol-1Molar mass of 20,000 g mol-1 solute = 20,000 g mol-1Molar concentration of the solution = (30/75,000) × 4 + (10/20,000) = 0.0008 mol L-1Step 5: Calculate the osmotic pressure using the formulaπ = nRT/Vwhereπ = osmotic pressuren = number of solute particlesR = gas constantT = temperatureV = volume of solutionπ = (3.9732 × 1019 × 8.314 × 293)/(1000 × 1000)= 7.46 × 105 PaTherefore, the osmotic pressure of the solution is 7.46 × 105 Pa.

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Answer 2

To calculate the osmotic pressure of the solution, the final answer the osmotic pressure of the solution is approximately [tex]0.045[/tex]atm. We can use the formula: π = (n/V)RT,

where π represents the osmotic pressure, n/V is the concentration of the solute in moles per liter, R is the ideal gas constant, and T is the temperature in Kelvin. First, let's calculate the concentration of each solute:

For the solute with a molecular weight of [tex]75,000[/tex], which completely dissociates into four solutes of equal size:

Concentration = [tex]30[/tex] g/L [tex]/ (75,000 g/mol)[/tex] [tex]= 0.0004 mol/L[/tex]

For the solute with a molecular weight of 20,000:

Concentration = 10 g/L / (20,000 g/mol) = 0.0005 mol/L

Now, let's calculate the total concentration of the solutes:

Total Concentration [tex]= 4 * (0.0004 mol/L) + 0.0005 mol/L[/tex] [tex]= 0.0019 mol/L[/tex]

Using the ideal gas constant R = 0.0821 L·atm/(mol·K) and assuming a typical room temperature of 298 K, we can calculate the osmotic pressure: π [tex]= (0.0019 mol/L) * (0.0821 L·atm/(mol·K)) * (298 K)[/tex]

[tex]= 0.045 atm[/tex]

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