Question

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?

Answer 1

Answer: 0.07 moles

Step-by-step 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
`p_(total)=p_(H_2O)+p_(H_2)`

Given:
`p_(total)=725torr`

`p_(H_2O)=24torr`

`p_(H_2)=?`

Thus
`725torr=24torr+p_(H_2)`

`p_(H_2)=701torr`

According to the ideal gas equation:'

`PV=nRT`

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

`n=(PV)/(RT)=(0.92* 1.85L)/(0.0821 * 308)=0.07moles`

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