Question

If water molecules pass through a membrane with a steady state flux of 220 mole/(m2 day), how long will it take, in hours, for 0.400 kg of water to pass through a 565 square centimeter of the membrane?

Answer 1

0.0386 hr

Step-by-step explanation:

Area = 565 cm^2 = 0.0565 m^2 (1 cm^2 = 0.0001 m^2)

flux state rate = 220 mole/m^2-day

There are 24 hrs in a day, therefore rate in hrs will be

220/24 = 9.17 mole/m^2-hr

mass of water = 0.4 kg

molar mass of water = (1 x 2) + 16 = 18 kg/mole

therefore,

mole of water = mass of water/molar mass of water

mole of water = 0.4/18 = 0.02 mole

mole flux = mole/area = 0.02/0.0565 = 0.354 mol/m^2

time that will be taken will be for water to pass = mole flux/mole flux rate

time = 0.354/9.17 = 0.0386 hr