Question

The student collects the H2(g) produced by the reaction and measures its volume over water at 298 K after carefully equalizing the water levels inside and outside the gas-collection tube, as shown in the diagram below. The volume is measured to be 45.6mL . The atmospheric pressure in the lab is measured as 765 torr , and the equilibrium vapor pressure of water at 298 K is 24 torr .(i) The pressure inside the tube due to the H2(g)

Answer 1

Answer:

The pressure of  H₂(g) = 741 torr

Explanation:

Given that:

The atmospheric pressure measured in the lab  = 765 torr

The vapor pressure of water = 24 torr

By applying Dalton's Law of Partial Pressure :

$P_{total} = P_{H_2}+P_{H_2O}$

Making The Pressure inside the tube due to the H₂(g) the subject of the formula :

we have:

$P_{H_2} = P_{total} - P_{H_2O}$

= (765 -24) torr

= 741 torr

Therefore; the pressure of  H₂(g) = 741 torr

Answer 2

The total pressure inside a container is equal to the sum of the pressure of the components of the gas mixture. The pressure inside the tube due to the Hydrogen gas is 741 torr.

Dalton's Law of Partial Pressure :

The total pressure inside a container is equal to the sum of the pressure of the components of the gas mixture.

The formula is

$\bold{\sum P = P_1 +P_2}$

Where,

$\bold{\sum P}$ - total pressure = 765 torr

P1 =  Pressure of water = 24 torr

P2 pressure of hydrogen = ?

Put the values and solve it for P2

$\bold{P_2 = (765 -24)}\\\\\bold{P_2 = 741}$

Therefore, the pressure inside the tube due to the Hydrogen gas is 741 torr.

To know more about Dalton's law of partial pressure.

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