Chemistry, 08.04.2020 04:35 ani69

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)

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Answer from: alyxkellar06

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:

Osmotic pressure of a solution is calculated by the formula,

Π= iMRT

Where,

Π is the osmotic pressure

is van't hoff factor

is molar concentration (mol/L)

is gas constant

is Temperature in Kelvin

Now,

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

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.

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.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

if you mean hydrogen and oxygen then water is the product

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