Question

1a. A mixture of hydrogen and xenon gases, in a 8.27 L flask at 27 °C, contains 0.932 grams of hydrogen and 22.2 grams of xenon. The partial pressure of xenon in the flask is  atm and the total pressure in the flask is  atm.

1b. A mixture of argon and methane gases is maintained in a 9.12 L flask at a pressure of 2.42 atm and a temperature of 10 °C. If the gas mixture contains 13.4 grams of argon, the number of grams of methane in the mixture is  g.

1c.

A mixture of hydrogen and xenon gases, at a total pressure of 950 mm Hg, contains 0.621 grams of hydrogen and 14.8 grams of xenon. What is the partial pressure of each gas in the mixture?

P_H2 =  mm Hg
P_Xe =  mm Hg

1d. A mixture of argon and methane gases contains argon at a partial pressure of 280 mm Hg and methane at a partial pressure of 510mm Hg. What is the mole fraction of each gas in the mixture?

X_Ar =
X_CH4 =

Answer 1

1a) Answer :-   

1)  

2)

Explanation :-

Given :-

Volume = V = 8.27 L

Mass of hydrogen = m = 0.932 g   Mass of xenon = m = 22.2 g

Temperature =T = 27 °C = 300 K

PXe =? PTotal = ?

We have  

Molar mass of Hydrogen (H2) = M = 2 g/mol

Molar mass of Xenon (Xe) = M = 131.3 g/mol

Now

we know that the ideal gass equation is

....(1)

where

p = pressure V = volume

T = temperature n = number of moles  

R = gas constant = 0.0821 L.atm/mol.K

but,  

  

thus equation (1) becomes

   ...(2)

i.e.  ....(30

Now we caculate partial pressure of xenon

i.e.  

  

Now we caculate partial pressure of hydrogen

  

i.e.   

The total pressure in the flask (PTotal) is

  

i.e.

i.e.   or

.

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1b)

Answer :-   

1) Mass of Methane = 9.8 g

Explanation :-

Given :-

Volume = V = 9.12 L

Mass of Argon = m = 13.4 g   

Temperature =T = 10 °C = 283 K

PTotal = 2.42 atm

Mass of Methane = m = ?

We have  

Molar mass ofmethane (CH4) = M = 16 g/mol

Molar mass of Argon (Ar) = M = 39.9 g/mol

Now

we know that the ideal gass equation is

....(1)

where

p = pressure V = volume

T = temperature n = number of moles  

R = gas constant = 0.0821 L.atm/mol.K

but,  

  

thus equation (1) becomes

   ...(2)

i.e.  ....(3)

Now we caculate partial pressure of Argon PAr =?   

therefore from equation (3) we have,

i.e.  

    

Now

we caculate partial pressure of methane

we know      

The total pressure in the flask (PTotal) is

  

i.e.   

i.e.

i.e.

Now

we calculate moles of methane

we have   

for methane

thus,

  

i.e.  

i.e.  

.

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1c)

Answer :-   

1)  

2)

Explanation :-

Given :-

PTotal = 950 mmHg

Mass of hydrogen = m = 0.621 g   Mass of xenon = m = 14.8 g

We have  

Molar mass of Hydrogen (H2) = M = 2 g/mol

Molar mass of Xenon (Xe) = M = 131.3 g/mol

First we calculate moles (n) of hydrogen and xenon

Formula :-   

Now

we calculate mole fraction (x) of hydrogen and xenon

Formula :-

thus,  

i.e.   

and

i.e.

Now

we caculate partial pressure of Xenon PXe and Hydrogen PH2

Formula :-

i.e.   

thus,

i.e.

and

.

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1d)

Answer :-   

1)  

2)  

Explanation :-

Given :-

   

We have  

Molar mass ofmethane (CH4) = M = 16 g/mol

Molar mass of Argon (Ar) = M = 39.9 g/mol

Now first we calculate total pressure of mixture (solution)

thus,

  

i.e.

Now,

we calculate mole fraction (x) of methan and argon

Formula :-

i.e.    

i.e.   

thus,

  

and

.  

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