Question

Calculate the osmotic pressure (in torr) of 6.00 L of an aqueous 0.958 M solution at 30.°C, if the solute concerned is totally ionized into three ions (e.g., it could be Na2SO4 or MgCl2).

Answer 1

The osmotic pressure of the 0.958 M solution is 23.96 atm.

Step-by-step explanation:

The osmotic pressure of a solution can be calculated using the formula II = MRT, where II is the osmotic pressure, M is the molarity of the solution, R is the gas constant (0.08206 L atm/mol K), and T is the temperature in Kelvin. In this case, we have a 0.958 M solution with a volume of 6.00 L at a temperature of 30 °C. To convert the temperature to Kelvin, we add 273 to get 303 K. Substituting the values into the formula, we get:

II = (0.958 M) (0.08206 L atm/mol K) (303 K)

II = 23.96 atm

Therefore, the osmotic pressure of the solution is 23.96 atm.

Answer 2

Answer: The osmotic pressure is 54307.94 Torr.

Step-by-step explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

`\pi=iCRT`

where,

`\pi` = osmotic pressure of the solution = ?

i = Van't hoff factor = 3

C = concentration of solute = 0.958 M

R = Gas constant =
`62.364\text{ L Torr }mol^(-1)K^(-1)`

T = temperature of the solution =
`30^oC=[30+273]K=303K`

Putting values in above equation, we get:

`\pi=3* 0.958mol/L* 62.364\text{ L. Torr }mol^(-1)K^(-1)* 303K\\\\\pi=54307.94Torr`

Hence, the osmotic pressure is 54307.94 Torr.