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Mass Concentration Problem

  • Thread starter Nimmy
  • Start date
  • #1
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Please check if right.

Determine the mass concentration for N20 at a mixing ratio of 311 ppb at P = 1 atm, T= 298 K.

My work

#density of N20 = Concentration of N20 * #density air

PV = #moles of air R T

PV/RT = # moles of air

PV/RT = Numberofmolecules/Avagardo # = #moles of air

PV*Avagardo#/RT = Number of molecules of air = # density of air * V

# density of air = P*Avagardo's #/RT

So...

#Densityof N20 = Concentration of N20* P*Avagardo #/RT

DensityofN20 = # density of N20* Molar mass N20/Avagardo's #

DensityofN20 = P*Avagardo's #*Concentration of N20/(RT *molar mass of N20/Avagardo's #)

Avagardo's # cancels out leaving:

Density of N20 = P*Concentration of N20*Molar Mass of N20/(RT)

Now plugging in the numbers lead to

Density of N20 = (1 atm)*(1.101325*10^5 Pa/atm)*(311*10^-9 mol/mol)*(46.00 g/mol) /((8.314 J/mol K)*298 K) = 0.0006 g/m^3

Is this correct step and answer? If so where is the error? Just needs to be verified. Thanks.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
157
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I tried to read your post several times but it is too annoying trying to follow the equations written out like that and I'm lazy.

I'd approach this by PV = NkT
take a billion particles of ideal gas, everything except volume is a given.
You are given 331 of these particles are N2O so just multiply 331 particles per volume by the mass per particle of N2O and you have your answer.

You might have done it right too there are several ways to calculate but I don't think you're going to get much of a response unless you enter your work via LaTex or scan in something hand written.
 
  • #3
41
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Much better now?

Let's try this.

Please check if right.

Determine the mass concentration for N20 at a mixing ratio of 311 ppb at P = 1 atm, T= 298 K.

My work

nd N20= CN20 * ndair

P*V = nmoles air* R * T

(P*V)/(RT) = nmoles air

(P*V)/(R*T) = N molecules air/Av = nmoles air

(P*V*Av)/(R*T) = N molecules air= ndair * V

ndair= (P*Av)/(RT)

So...

ndN20 = CN20* (P*Av/RT)

ρN20 = ndN20* (MN20/Av)

ρN20= ((P*Av*CN20)/(RT *MN20/Av))

Avagardo's # cancels out leaving:

ρN20= (P*CN20*MN20/(R*T))

Now plugging in the numbers lead to

ρN20= (1 atm)*(1.101325*10^5 Pa/atm)*(311*10^-9 mol/mol)*(46.00 g/mol) /((8.314 J/mol K)*298 K) = 0.0006 g/m^3

Is this correct step and answer? If so where is the error? Just needs to be verified. Thanks.
 
  • #4
20,143
4,212
Find the moles/liter of air (p/RT). Multiply by the mixing ratio to get the moles/liter of N2O. Multiply by the molecular weight of N2O to get the density of N2O.
 
  • #5
41
0
Find the moles/liter of air (p/RT). Multiply by the mixing ratio to get the moles/liter of N2O. Multiply by the molecular weight of N2O to get the density of N2O.
Perfect that matches with what I wrote it seems. Thanks. :)
 

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