Question

You have a 0.2 molar concentration of sugar solution in an open beaker.

Calculate the water potential at 22 degrees Celsius.
Make me happy: show your units cancelling, just this once. You should be left with bars.

Answer 1

The water potential at 22 degrees Celsius for a 0.2 Molar sugar solution in an open beaker is approximately -4.876 bars.

Step-by-step explanation:

The water potential of a solution can be determined using the formula:

Ψ = Ψs + Ψp

where Ψs represents the solute potential, and Ψp represents the pressure potential.

The solute potential (Ψs) can be calculated using the formula:

Ψs = -iCRT

In this case, we can assume that the water potential of pure water (Ψ) is 0 bars. So, we can write the equation as:

0 = Ψs + Ψp

Since the solution is in an open beaker, the pressure potential (Ψp) can be assumed to be 0 bars.

Therefore, the water potential (Ψ) is equal to the solute potential (Ψs).

Given the concentration of the sugar solution is 0.2 Molar, we need to find the solute potential.

Plugging the values into the formula:

Ψs = -iCRT

We know that:

i = van 't Hoff factor (number of particles into which the solute dissociates)

C = molar concentration (0.2 Molar)

R = ideal gas constant (0.0831 liter bar / mol K)

T = temperature in Kelvin (22°C + 273 = 295 K)

If sugar is a non-electrolyte, it does not dissociate into ions and the van 't Hoff factor for sugar (C12H22O11) is 1.

Plugging in the values and solving for Ψs:

Ψs = -(1)(0.2)(0.0831)(295)

Ψs ≈ -4.876 bars

Therefore, the water potential at 22°C for the 0.2 Molar sugar solution in the open beaker is approximately -4.876 bars.

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