To solve this problem, we need to use the formula for osmotic pressure:

Π = nRT/V

where:
Π is the osmotic pressure,
n is the number of moles of solute,
R is the ideal gas constant,
T is the temperature in Kelvin,
V is the volume of the solution in liters.

First, we need to calculate the number of moles of KCl. The molar mass of KCl is approximately 74.55 g/mol. So,

n = 10.0 g / 74.55 g/mol = 0.134 mol

The ideal gas constant R is 0.0821 L·atm/(K·mol) and the temperature T is given as 298 K. The volume V is 1.00 L.

Substituting these values into the formula gives:

Π = (0.134 mol * 0.0821 L·atm/(K·mol) * 298 K) / 1.00 L = 3.28 atm

However, the options are given in kPa. We know that 1 atm is approximately equal to 101.325 kPa, so we need to convert:

Π = 3.28 atm * 101.325 kPa/atm = 332 kPa

So, the osmotic pressure of the solution is approximately 332 kPa.

Question

To solve this problem, we need to use the formula for osmotic pressure:

Π = nRT/V

where:
Π is the osmotic pressure,
n is the number of moles of solute,
R is the ideal gas constant,
T is the temperature in Kelvin,
V is the volume of the solution in liters.

First, we need to calculate the number of moles of KCl. The molar mass of KCl is approximately 74.55 g/mol. So,

n = 10.0 g / 74.55 g/mol = 0.134 mol

The ideal gas constant R is 0.0821 L·atm/(K·mol) and the temperature T is given as 298 K. The volume V is 1.00 L.

Substituting these values into the formula gives:

Π = (0.134 mol * 0.0821 L·atm/(K·mol) * 298 K) / 1.00 L = 3.28 atm

However, the options are given in kPa. We know that 1 atm is approximately equal to 101.325 kPa, so we need to convert:

Π = 3.28 atm * 101.325 kPa/atm = 332 kPa

So, the osmotic pressure of the solution is approximately 332 kPa.

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for osmotic pressure:

Π = nRT/V

where:
Π is the osmotic pressure,
n is the number of moles of solute,
R is the ideal gas constant,
T is the temperature in Kelvin,
V is the volume of the solution in liters.

First, we need to calculate the number of moles of KCl. The molar mass of KCl is approximately 74.55 g/mol. So,

n = 10.0 g / 74.55 g/mol = 0.134 mol

The ideal gas constant R is 0.0821 L·atm/(K·mol) and the temperature T is given as 298 K. The volume V is 1.00 L.

Substituting these values into the formula gives:

Π = (0.134 mol * 0.0821 L·atm/(K·mol) * 298 K) / 1.00 L = 3.28 atm

However, the options are given in kPa. We know that 1 atm is approximately equal to 101.325 kPa, so we need to convert:

Π = 3.28 atm * 101.325 kPa/atm = 332 kPa

So, the osmotic pressure of the solution is approximately 332 kPa.

What is the osmotic pressure of a solution prepared from 10.0 g of theelectrolyte KCl and enough water to make 1.00 L of solution at a temperature of 298K?Question 8Select one:665 kPa332 kPa27.9 kPa166 kPa55.8 kPa

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