Question

The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 torr and the vapor pressure of water under the experimental conditions is 23.8 torr, what is the pressure (torr) of the H 2 gas

Answer 1

Answer:

the pressure (torr) of theH₂ gas is 736.2 torr

Explanation:

Given that

The H₂ gas  produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.

So; the total pressure in the eudiometer = Pressure in the H₂ gas  - Pressure of the water vapor

Where;

$P_{Totsl}$ =  total pressure in the eudiometer = 760.0 torr

$P_{H_2}$ = Pressure in the H₂ gas  = ???

$P_{H_2O}$ = Pressure in the water vapor  = 23.8 torr

Now:

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

- $P_{H_2}$ = + $P_{H_2O}$ - $P_{Totsl}$

$P_{H_2}$  = - $P_{H_2O}$ + $P_{Totsl}$

$P_{H_2}$  =  (- 23.8 + 760) torr

$P_{H_2}$  = 736.2 torr

Thus; the pressure (torr) of theH₂ gas is 736.2 torr