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 heptahydrate (MgSO4·7H2O) form, also known as Epsom salt. The densities of the anhydrate and heptahydrate crystalline forms are 2.66 and 1.68 g/mL, respectively. Suppose you wish to form a 20.0 wt% MgSO4 aqueous solution by simply pouring crystals of one of the forms into a tank of water while the temperature is held constant at 30°C. The specific gravity of the 20.0 wt% solution at 30°C is 1.22. Answer the following questions for both forms of the MgSO4 crystals: What volume of water should be in the tank before crystals are added if the final product is to be 1000 kg of the 20 wt% 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?

Answer 1

Answer:

Magnesium sulphate is in two forms

Magnesium anhydrate and magnesium hepta hydrate

Density of MgSO4= 2.66 g/ml

Density of MgSO4. 7H2O= 1.68 g/ml

M. W of MgSO4= 120 kg/kmol

In order to form 1000 Kg 20wt% solution

A) the Kg of water required before addition of crystals for MgSO4 =(1000) (0.80)= 800 kg

Average density= (0.20) (2.66) +(0.80) (1) = 1.332 g/ml = 1332 Kg/m3

Volume of water required= (800/1332) = 600.60 litres

For hepta hydrate

Average density= (0.20) (1.68) +(0.80) (1) = 1.136 g/ml=

1136 kg/m 3

MgSO4. 7 H2O, mole fraction of water present =

7×18/(120+(7×18) )=0.512 = 51.2 mol%= 13.6 wt%

1000 kg of (20 wt%) solution required

Amount of pure MgSO4 needed= 1000(.20) = 200 Kg

But it contains 13.6% water hence total weight of

MgSO4. 7H2O= 200/(1-0.136) = 231.48 kg

Amount of water required= (1000-231.48) = 768.51 Kg

Volume of water required= 768.51/1136= 676.51 Litres

B) diameter of tank= 0.30 m

V= 3.142(d2)(H) /4

For pure MgSO4 , volume of water required=600.60 litres

H = 8.49 m

For hepta hydrate volume of water required=676.51 litres

H= 9.57 m

C) height of tank after addition of crystals

Total volume in case of pure MgSO4=

1000/(1332) = 0.7507= 750.7 litres

H= 10.62 m

Total volume of hepta hydrate =

1000/(1136) = 0.8802= 880.2 litres

H= 12.45 m

D) after dissolution of crystals

The volume after dissolution in case of pure MgSO4

= volume of water= 600.67 litres

H=8.49 m

in case of hepta hydrate

The volume after dissolution= ( volume of water + volume of water in MgSO4. 7H2O)

= (676.51+(231.48×0.136) = 707.99 Litres

H= 10.01 m

Part B question

A) Weight of house= 1×105 lb= 2.20 ×105

Force applied by house= (2.20×105) (9.81) = 2.15×106 N

Pressure of ballon= 1.05 atm= 1.063×105 Pa

Diameter= 9.5 inch= 0.2413 m

Area= (3.142) (0.2413) 2/4= 0.0457 m2

F= P×A= 4857.91 N

To float the house

Both forces should be equal

So force applied by 1 ballon= 4857.91 N

Total force required=2.15×106 N

Number of ballons required= (2.15×106) /4857.91=442.57=443 ballons

B) the ballon pressure must be greater than atmospheric pressure

If both pressures are equal then there will be no air flow

If outside pressure is higher, then air would flow from outside to inside of ballon causing deflation, hence pressure inside the ballon must be higher than outside