Question

Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air and others based on the high solubility of the heptahydratc (MgSO.4 . 7H2,O)form, also known as Epsom salt. The densities of the anhydrate and heptahydratc crystalline forms are 2.66 and 1.68 g/mL, respectively. Suppose you wish to form a 20.0 wt% MgS()4 aqueous solution by simply pouring crystals of one of the forms into a tank of water while the temperature is held constant at 30degree C. The specific gravity of the 20.0 wt% solution at 30degree C is 1.22. Answer the following questions for both forms of the MgS04 crystals: What volume of water should be in the tank before crystals arc added if the final product is to be 1000 kg of the 20wt% solution? Suppose the tank diameter is 0.30 m. What is the height of liquid in the tank before the crystals are added? What is the height of the water in the tank after addition of the crystals but before they begin to dissolve? What is the height of liquid in the tank after all the MgSO4 has dissolved?5.8. Use the idcal-gas equation of state to kg/m3 at 40degree C

Answer 1

a) the final product is to be 1000Kg of the 20wt% solution

20wt% solutio means = 20grams of Magnesium in 100g of solution

so in 1000Kg of solution the amount of salt needed = 200Kg

so for each salt we will add 200Kg of salt in the tank and 800Kg of water

b) volume occupied by water before adding salt = 800L

As the density of water = 1Kg / L

the radius of tank = 0.30/2 m = 0.15 m

volume of tank = πr²h = 800L = 800dm^3 = 0.8m^3

0.8m^3 = πr²h = 3.14 X (0.15)^2 x height

height = 11.32 meters

c) density of crystals will play the role now

(i) density of MgSO4 crystals = 2.66Kg / L

volume occupied by 200Kg of MgSO4 = mass / Density = 200 / 2.66 = 75.19 Litres = 0.07519m^3

total volume by water + MgSO4 = 0.07519 + 0.8 m^3 = 0.875 m^3

0.875m^3 = πr²h = 3.14 X (0.15)^2 x height

height = 12.38 meters

(ii) for hydrate

density of MgSO4 hydrates crystals = 1.68Kg / L

volume occupied by 200Kg of MgSO4 = mass / Density = 200 / 1.68 = 119.05 Litres = 0.119m^3

total volume by water + MgSO4 = 0.119 + 0.8 m^3 =0.919m^3

0.9195m^3 = πr²h = 3.14 X (0.15)^2 x height

height = 13.01meters

c) after dissolution of MgSO4

density of solution =1.22

Volume of solution =Mass / density = 1000 / 1.22 = 819.67 Litres = 0.8197 m^3

0.8197m^3 = πr²h = 3.14 X (0.15)^2 x height

height = 11.60 meters