Question

In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water contains about 560 mg of dissolved solids per milliliter. Reverse osmosis units are marketed in this area to soften water. A typical unit exerts a pressure of 8.0 atm and can produce 45 L water per day
a. Assuming all of the dissolved solids are Mgco3 and assuming a temperature of 27 degrees, what total volume of water must be processed to produce 45 L pure water?
b. Would the same system work for purifying seawater? (Assume seawater is 0.60 MNaCl)

Answer 1

Answer:

a) $46.734L$ of hard water is require to produce $45L$ 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=$iMRT$

Where,

Π is the osmotic pressure

$i$ is van't hoff factor

$M$ is molar concentration (mol/L)

$R$ is gas constant

$T$ is Temperature in Kelvin

Now,

$MgCO_{3}\rightarrow Mg^{2+}+ CO_{3}^{2-}$

So, according to the above equation $i=2$

Calculation concentration at which reverse osmosis will stop

$M=\frac{ Π }{iRT}$

$M=\frac{8atm}{2\times0.082L atm mol^{-1}K^{-1} \times 300K} \\M= 0.162M$

Calculating Initial concentration that is moles of $MgCO_{3}$

$n=\frac{given mass}{molar mass} \\\\n=\frac{560 \times 10^{-6}g}{84.32g/mol}\\\\N=6.642\times10^{-6}moles$

Calculating molarity

$M= 6.642 \times 10^{-6}\times 10{^3}\\M=6.642\times10^{-3}M$

Total number of moles in total volume must remain the same,

$C_{1}V_{1}=C_{2}V_{2}$

$0.006642M \times V_{1}=0.162M \times V_{2}$

$\frac{V_{1}}{V_{2}}= \frac{0.162}{0.006.642}=25.953$

$V_{1}=25.953 V_{2}$

Also, $V_{1}=45L$

$V_{2}=\frac{45}{25.953}\\ V_{2}= 1.734L$

So, net Volume,

$V=45L+1.734L=46.734L$

So, $46.734L$ is require to produce $45L$ of pure water.

b)

Using the equation for $0.60M NaCl$ as well,

$C_{1}V_{1}=C_{2}V_{2}$

$C_{1}=0.60M\\V_{1}=45L\\C_{2}=0.162M\\V_{2}=?$

$V_{2}=\frac{0.60 \times 45}{0.167}\\V_{2}= 166.66L$

$Net water=166.66+45L=211.66L$

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.