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The allowable concentration level of Vinyl Chloride C_2 H_3 Cl, in the atmosphere in a...

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spock4 | Student, Undergraduate | Honors

Posted June 27, 2012 at 1:52 PM via web

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The allowable concentration level of Vinyl Chloride C_2 H_3 Cl, in the atmosphere in a chemical plant is 2.05x10^-6 g/L. How many moles of Vinyl...

...Chloride in each liter does this represent? How many molecules per liter?

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chaobas | College Teacher | Valedictorian

Posted June 28, 2012 at 1:33 AM (Answer #1)

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now 

density of 

vinly chloride = 2.05*10^-6 g/L

So if we take 1 L of the soltuion, the mass of vinyl chloride would be 2.05*10^-6 g

as density = mass/volume

mass = density * volume

Mass of vinly chloride = 2.05*10^-6 g/L) * 1 L = 2.05*10^-6 g

and the molar mass is 62.50 g/mol (C2H3Cl = 6*2 + 1*3+35.45*1 = 62.50 )

 

Moles = mass/molar mass

Moles of vinyl = (2.05*10^-6 g )/(62.5 g /mol) = 3.28*10^-8 mol

 

So in 1 L there would be 3.28*10^-8 mol of vinyl chloride.

So it is 3.28*10^-8 mol/L of vinyl chloride

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sanjeetmanna | College Teacher | (Level 3) Assistant Educator

Posted August 21, 2012 at 2:08 AM (Answer #2)

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Given 

`2.05*10^-6 g/L` of vinyl chloride, which means `2.05*10^-6` grams of vinyl chloride present in 1 litre.

Moles = `(mass)/(molar mass)`

Moles = `(2.05*10^-6)/(62.498)`

Moles = `3.28*10^-8`

So in 1 litre it represent `3.28*10^-8` vinyl chloride.

According to avogadro number...

1 mole = avogadro number number of molecules 

That is...

1 mole = `6.022*10^23` molecules.

Therefore `3.28*10^-8` moles of vinyl chloride will contain...

`(3.28*10^-8) * (6.022*10^23) = 19.75216*10^15` molecules.

Therefore `2.05*10^-6` `g/L` of vinyl chloride represent `3.28*10^-8` moles with `19.75216*10^15` molecules.

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