Consider the following data for air trapped in a flask: Pressure = 0.988 atm Room Temperature = 23.5°C Volume of the flask = 1.042 L For this calculation, assume air is 78.5% nitrogen and 21.5% oxygen (by number of moles). R = 0.0821 L•atm•mol-1•K-1 N = 14.01 g/mol O = 16.00 g/mol What is the mass of air in the flask

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Answer:

total mass will be =  = 1.207g

Explanation:

First what is given  

Pressure P= 0.988 atm                  Room TemperatureT = 23.5°C= 296.5 K

Volume V= 1.042 L

Nitrogen in air is 80 % (moles number) = 0.8                

Ideal gas constant R = 0.0821 L atmmol-1K-1

Given mass of N = 14.01 g/mol

Given mass of oxygen O = 16.00 g/mol

Total number of moles = ?

So first we have to find the total number of moles by using formula  

Total number of moles  

n = PV/ RT  

adding the values  

moles n= 0.988atm x 1.042L / (0.0821L-atm/mole-K x 296.6K)  

  = 1.0294 / 24.35

= 0.042 moles (total number of moles)

So by using Nitrogen percentage  

Moles of nitrogen = total moles x 80/100

                            = 0.042moles x 0.8  

         

So moles of O2= Total moles – moles of N2  

                       =   0.042moles - 0.034moles

    Moles of O2 = 0.008moles

Now for finding the mass of the N2 and oxygen  

Mass of Nitrogen N2 = no of moles x molar mass

                           = 0.034 x 28

                           = 0.952 g

Mass of oxygen O2 = no of moles x molar mass

                           = 0.008 x 32

                           =  0.256 g

total mass will be = Mass of Nitrogen N2 + Mass of oxygen O2  

                             =0.952 g  + 0.256 g

                               =1.207g

The mass of air in the flask = 1.207g

Given:

Pressure, P= 0.988 atm                  

Room Temperature, T = 23.5°C= 296.5 K

Volume, V= 1.042 L

Nitrogen in air is 80 % (moles number) = 0.8                

Ideal gas constant R = 0.0821 L atm [tex]mol^{-1}K^{-1}[/tex]

Molar mass of N = 14.01 g/mol

Molar mass of oxygen O = 16.00 g/mol

To find:

Total number of moles = ?

Calculation for number of moles:

From ideal gas law:

n = PV/ RT  

n= 0.988atm * 1.042L / (0.0821 L atm [tex]mol^{-1}K^{-1}[/tex] * 296.6K)  

n  = 1.0294 / 24.35

n= 0.042 moles

Using mol fraction we will calculate moles for nitrogen and oxygen:

Moles of nitrogen = total moles * 80/100

Moles of nitrogen = 0.042moles * 0.8 = 0.034 moles

So, Moles of O₂ = Total moles – moles of N₂  

Moles of O₂ =   0.042 moles - 0.034 moles

Moles of O₂ = 0.008 moles

Calculation for mass:

Mass of Nitrogen N₂ = no of moles x molar mass

= 0.034 x 28

= 0.952 g

Mass of oxygen O₂ = no of moles x molar mass

= 0.008 * 32

=  0.256 g

Total mass will be = Mass of Nitrogen N₂ + Mass of oxygen O₂

=0.952 g  + 0.256 g

Total mass = 1.207g

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