The solution before dilution and after dilution contains same number of moles, and water is added for dilution.
Option B
<h3><u>Explanation:</u></h3>
Suppose before dilution, the solution contains x moles of KCl in Y liter of water. Now as the concentration got halved, then the solution contains x moles of KCl in 2Y kiters of solution. So the number of moles of KCl in the solution remained constant.
Again, as the solution is diluted to half of the concentration, water must have been added with the solution to make it dilute.
The formula for the change in Gibbs energy of a solid is:
ΔG = Vm ΔP
where, ΔG is change in Gibbs, Vm is molar volume, ΔP is
change in pressure
ΔP = P(final) – P(initial)
P(final) = 1 atm = 101325 Pa
P(initial) = ρ_water *g *h = (1030 kg/m^3) * 9.8 m/s^2 *
2000 m = 20188000 kg m/s^2 = 20188000 Pa
Vm = (950 kg/m^3) * (1000 mol / 891.48 kg) = 1065.64
mol/m^3
So,
ΔG = (1065.64 mol/m^3) * (101325 Pa - 20188000 Pa)
<span>ΔG = -21405164347 J = -21.4 GJ</span>
The surrounding ecosystem in and around the water