# The mole fraction of argon in a gas mixture, which contains only argon and hydrogan, is0.650. Calculate the density of the gas mixture at a pressure of 1.50 bar and a temperature of3.00 × 102 K....

The mole fraction of argon in a gas mixture, which contains only argon and hydrogan, is

0.650. Calculate the density of the gas mixture at a pressure of 1.50 bar and a temperature of

3.00 × 102 K. State the assumption you used in this calculation.Relative atomic masses:Ar =40,H =1

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### 1 Answer

Here we can use ideal gas equation for the calculation.

`PV= nRT`

`n = (PV)/(RT)`

Where;

P = Pressure

V = Volume

n = Moles

R = Universal gas constant

T = temperature

Also we know that;

`P_A = X_AP_T`

Where;

`P_A` = Partial pressure of A

`X_A` = mole fraction of A

`P_T ` = Total pressure

Assume the volume to be `1m^3` .

`1 bar = 1xx10^5pa`

For Ar;

`n = (0.65xx1.5xx10^5)/(8.314xx300)`

`n = 39.09mol`

For `H_2` ;

`n = (0.35xx1.5xx10^5)/(8.314xx300)`

`n = 21.05mol`

Mass of Ar `= 39.09xx40 = 1593.6`

Mass of `H_2` `= 21.05xx2 = 42.1`

Total mass `= 1593.6+42.1 = 1635.7g = 1.64kg`

Total Volume `= 1m^3`

*Density of the mixture = 1.64kg/m^3*

*Assumption;*

*Gasses behave as ideal gasses**Gasses don't react with each other**No internal bonds between gas molecules or all bonds are equal*

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