Respuesta :
Answer:
a) [tex]46.734L[/tex] of hard water is require to produce [tex]45L[/tex] of pure water.
b) Same water would not work for purifying sea water as it would require so much of energy.
Explanation:
a)
Osmotic pressure of a solution is calculated by the formula,
Π=[tex]iMRT[/tex]
Where,
Π is the osmotic pressure
[tex]i[/tex] is van't hoff factor
[tex]M[/tex] is molar concentration (mol/L)
[tex]R[/tex] is gas constant
[tex]T[/tex] is Temperature in Kelvin
Now,
[tex]MgCO_{3}\rightarrow Mg^{2+}+ CO_{3}^{2-}[/tex]
So, according to the above equation [tex]i=2[/tex]
Calculation concentration at which reverse osmosis will stop
[tex]M=\frac{ Π }{iRT}[/tex]
[tex]M=\frac{8atm}{2\times0.082L atm mol^{-1}K^{-1} \times 300K} \\M= 0.162M[/tex]
Calculating Initial concentration that is moles of [tex]MgCO_{3}[/tex]
[tex]n=\frac{given mass}{molar mass} \\\\n=\frac{560 \times 10^{-6}g}{84.32g/mol}\\\\N=6.642\times10^{-6}moles[/tex]
Calculating molarity
[tex]M= 6.642 \times 10^{-6}\times 10{^3}\\M=6.642\times10^{-3}M[/tex]
Total number of moles in total volume must remain the same,
[tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
[tex]0.006642M \times V_{1}=0.162M \times V_{2}[/tex]
[tex]\frac{V_{1}}{V_{2}}= \frac{0.162}{0.006.642}=25.953[/tex]
[tex]V_{1}=25.953 V_{2}[/tex]
Also, [tex]V_{1}=45L[/tex]
[tex]V_{2}=\frac{45}{25.953}\\ V_{2}= 1.734L[/tex]
So, net Volume,
[tex]V=45L+1.734L=46.734L[/tex]
So, [tex]46.734L[/tex] is require to produce [tex]45L[/tex] of pure water.
b)
Using the equation for [tex]0.60M NaCl[/tex] as well,
[tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
[tex]C_{1}=0.60M\\V_{1}=45L\\C_{2}=0.162M\\V_{2}=?[/tex]
[tex]V_{2}=\frac{0.60 \times 45}{0.167}\\V_{2}= 166.66L[/tex]
[tex]Net water=166.66+45L=211.66L[/tex]
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.
