What is the water potential of a 5.7 L water in which 2.1 moles of NaCl has been dissolved at 30 degrees celcisus?

Help please? Can you show work as well?
It'll be greatly appreciated. Thanks in advance.

To find the water potential of the solution, you will need to use the formula:

Ψ = Ψs + Ψp

where:
- Ψ is the water potential
- Ψs is the solute potential
- Ψp is the pressure potential

To calculate the solute potential (Ψs), you can use the formula:

Ψs = -iCRT

where:
- i is the ionization constant (the number of particles each solute molecule splits into when dissolved)
- C is the molar concentration of the solute (in moles per liter)
- R is the gas constant (0.0831 liter bar per mole kelvin)
- T is the temperature in kelvin (30°C = 303 K)

In this case, you have 2.1 moles of NaCl dissolved in 5.7 L of water:

- i = 2 (NaCl particles dissociate into two ions: Na+ and Cl-)
- C = 2.1 moles / 5.7 L = 0.368 moles per liter
- R = 0.0831 liter bar per mole kelvin
- T = 30°C + 273 = 303 K

Plugging these values into the formula, you get:

Ψs = -(2)(0.368)(0.0831)(303)
= -18.11 bar

Next, you need to calculate the pressure potential (Ψp). Assuming the system is open to the atmosphere, the pressure potential is normally zero unless stated otherwise.

Ψp = 0

Finally, you can find the overall water potential (Ψ) by summing the solute potential (Ψs) and the pressure potential (Ψp):

Ψ = Ψs + Ψp
= -18.11 bar + 0
= -18.11 bar

Therefore, the water potential of the solution is -18.11 bar.