Answer:
a)
of hard water is require to produce
of pure water.
b) Same water would not work for purifying sea water as it would require so much of energy.
Step-by-step explanation:
a)
Osmotic pressure of a solution is calculated by the formula,
Π=
![iMRT](https://img.qammunity.org/2021/formulas/chemistry/high-school/srku96mg0cwy9bt3f4jtv7sir51y3qrrkh.png)
Where,
Π is the osmotic pressure
is van't hoff factor
is molar concentration (mol/L)
is gas constant
is Temperature in Kelvin
Now,
![MgCO_(3)\rightarrow Mg^(2+)+ CO_(3)^(2-)](https://img.qammunity.org/2021/formulas/chemistry/high-school/jresvuwdni26mmxvxmxq2cm2cffk87byzt.png)
So, according to the above equation
![i=2](https://img.qammunity.org/2021/formulas/chemistry/high-school/r4a7lb5nd3g368i9rwerldlnm0htds1l51.png)
Calculation concentration at which reverse osmosis will stop
![M=( Π )/(iRT)](https://img.qammunity.org/2021/formulas/chemistry/high-school/7kpfug4en5me9pcoz05huo41zy7ujm9huy.png)
![M=(8atm)/(2*0.082L atm mol^(-1)K^(-1) * 300K) \\M= 0.162M](https://img.qammunity.org/2021/formulas/chemistry/high-school/53yu4pmuupfrpp0c4tfgot9zz5varaixb9.png)
Calculating Initial concentration that is moles of
![MgCO_(3)](https://img.qammunity.org/2021/formulas/chemistry/high-school/m9lzggadu3iez8zboah2dfyopz14pt6etd.png)
![n=(given mass)/(molar mass) \\\\n=(560 * 10^(-6)g)/(84.32g/mol)\\\\N=6.642*10^(-6)moles](https://img.qammunity.org/2021/formulas/chemistry/high-school/2hvcgrn803vrokszvc63fn654b4m4odlas.png)
Calculating molarity
![M= 6.642 * 10^(-6)* 10{^3}\\M=6.642*10^(-3)M](https://img.qammunity.org/2021/formulas/chemistry/high-school/76taer8uxnnpstgr2xmaoxx0e35q4l9tfq.png)
Total number of moles in total volume must remain the same,
![C_(1)V_(1)=C_(2)V_(2)](https://img.qammunity.org/2021/formulas/chemistry/high-school/w0gbe1xu6079mgaq2dwd22847epdtizw3n.png)
![0.006642M * V_(1)=0.162M * V_(2)](https://img.qammunity.org/2021/formulas/chemistry/high-school/qag91bt3c22my4erxn4ik9svl5pftgnce5.png)
![(V_(1))/(V_(2))= (0.162)/(0.006.642)=25.953](https://img.qammunity.org/2021/formulas/chemistry/high-school/49nvhhg9mp2ekgkx1ioxy7ddlh6acd29l3.png)
![V_(1)=25.953 V_(2)](https://img.qammunity.org/2021/formulas/chemistry/high-school/l9e4p4l8auqg5mun867s0uw0r03x7yzylj.png)
Also,
![V_(1)=45L](https://img.qammunity.org/2021/formulas/chemistry/high-school/lzm4vscf8e4tbam510kjzau6euok62dmmq.png)
![V_(2)=(45)/(25.953)\\ V_(2)= 1.734L](https://img.qammunity.org/2021/formulas/chemistry/high-school/l9ofgbcy9kuojtiqcwfi93z5tbo8vziiyx.png)
So, net Volume,
![V=45L+1.734L=46.734L](https://img.qammunity.org/2021/formulas/chemistry/high-school/iejyjxtsqxov9xls54f9gu0fx77xv5v35e.png)
So,
is require to produce
of pure water.
b)
Using the equation for
as well,
![C_(1)V_(1)=C_(2)V_(2)](https://img.qammunity.org/2021/formulas/chemistry/high-school/w0gbe1xu6079mgaq2dwd22847epdtizw3n.png)
![C_(1)=0.60M\\V_(1)=45L\\C_(2)=0.162M\\V_(2)=?](https://img.qammunity.org/2021/formulas/chemistry/high-school/dtgwwz9hmwbfqpv96jznnirdr38qq7rspq.png)
![V_(2)=(0.60 * 45)/(0.167)\\V_(2)= 166.66L](https://img.qammunity.org/2021/formulas/chemistry/high-school/g64iidd0abvf91p95l6zpzs3ri08ynh54e.png)
![Net water=166.66+45L=211.66L](https://img.qammunity.org/2021/formulas/chemistry/high-school/rlje8c39wje1hu9vkp494cym2tzrlb5z3q.png)
So, the same water would not work for purifying sea water as it would require so much of energy, as we require 211.66L of water to produce 45L of water.