7. A sample of hydrogen gas is mixed with water vapor. The mix has a total pressure of 725 torr and the water vapor has a partial pressure of 24 torr. What amount (in moles) of hydrogen gas is contained in 1.85 L of this mixture at 308 K?

Respuesta :

Answer: 0.07 moles

Explanation:

According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.

Thus [tex]p_{total}=p_{H_2O}+p_{H_2}[/tex]

Given: [tex]p_{total}=725torr[/tex]

[tex]p_{H_2O}=24torr[/tex]

[tex]p_{H_2}=?[/tex]

Thus [tex]725torr=24torr+p_{H_2}[/tex]

[tex]p_{H_2}=701torr[/tex]

According to the ideal gas equation:'

[tex]PV=nRT[/tex]

P= Pressure of the gas = 701 torr = 0.92 atm     (760torr=1atm)

V= Volume of the gas = 1.85 L

T= Temperature of the gas = 308 K

R= Value of gas constant = 0.082 Latm\K mol

[tex]n=\frac{PV}{RT}=\frac{0.92\times 1.85L}{0.0821 \times 308}=0.07moles[/tex]

Thus 0.07 moles of hydrogen gas is contained in 1.85 L of this mixture at 308 K.

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