[Saionji]

Gypsum is formed by CaSO4. Suppose there is a crack in the ceiling, through which drain water at a flow rate b = 2.0 dm ³ / day. If the plaster ceiling that has e = 1.50 cm thick, how long would it take to open a circular hole of diameter d = 1 cm?
Assuming that the density of plaster is d = 0.97 g/cm3 and its solubility product constant of KPS = 2 10-4 and 120g/mol= M (molar mass of CaSO4).

[eleven6eleven]

First we find the volume of plaster required to dissolve:

1.50 cm * ((1 cm)/2)^2 * pi = 1.18 cm^3

Now we multiply that by the density 0.97 g/cm^3:

1.18 cm^3 * 0.97 g/cm^3 = 1.14 g

Now we divide by the molar mass of gypsum, or calcium sulfate:

1.14 g/(120 g/mol) = 0.0095 mol CaSO4

So we know that ksp = [Ca2+][SO4 2-] = 2 x 10^-4.
In order for 0.0095 mol CaSO4 to dissolve in water, we have to figure out how much CaSO4 dissolve in 1 L of water.

x * x = 2 x 10^-4
x = 0.014 mol / L

So we now know the amount of water needed to dissolve 0.0095 mol CaSO4:

0.0095 mol / (0.014 mol / L) = 0.67 L

And since a decimeter cubed is a liter, 2 L flows per day.
So it would take this long:

0.67 L / (2 L / day) = 0.34 day = 8.1 hours.

Hope this helps! Let me know if I made a mistake.

[Borek]

Let me know if I made a mistake.

Yes, you did a mistake. Please read forum rules, especially To Replyers section.

[eleven6eleven]

Oops sorry...