Question

A(n) 564 mL sample of hydrogen was col lected over water at 21. 0°C on a day when the batometric pressure was 757 torr. What volume would the dry hydrogen occupy under standard conditions? The vapor pressure of water at 21. 0°C is 19. 0 torr. Answer in units of mL

Answer 1

The dry hydrogen volume under standard conditions is obtained by subtracting the vapor pressure of water from the barometric pressure to get the dry hydrogen pressure, then using the combined gas law formula to solve for the new volume at STP.

Step-by-step explanation:

The student has a chemistry problem involving the collection of hydrogen gas over water. To find the volume of the dry hydrogen gas at standard temperature and pressure (STP), one must first subtract the vapor pressure of water from the barometric pressure, and then employ the combined gas law.

The total pressure exerted by the gases is the sum of the partial pressures (Dalton's Law). The pressure of dry hydrogen is equal to the barometric pressure minus the vapour pressure of water. Thus, the pressure of the dry hydrogen is 757 torr - 19 torr = 738 torr.

Next, we apply the combined gas law:
P1V1/T1 = P2V2/T2

Here, P1 = 738 torr, V1 = 564 mL, T1 = 294.15 K (21°C + 273.15), P2 = standard pressure (760 torr), and T2 = standard temperature (273.15 K).

Rearranging and solving for V2:
V2 = P1V1T2 / (P2T1)

V2 = (738 torr * 564 mL * 273.15 K) / (760 torr * 294.15 K)

After calculating, we find the volume of dry hydrogen under STP conditions.

Answer 2

To find the volume of dry hydrogen gas under standard conditions, we need to consider the partial pressure of the hydrogen gas and the vapor pressure of water at the given temperature. We can use Dalton's law of partial pressures to calculate the partial pressure of the hydrogen gas, and then use the ideal gas law to find the volume of dry hydrogen gas under standard conditions. The volume of the dry hydrogen gas under standard conditions is approximately 548 mL.

Step-by-step explanation:

To find the volume of dry hydrogen gas under standard conditions, we need to consider the partial pressure of the hydrogen gas and the vapor pressure of water at the given temperature.

The total pressure is the sum of the partial pressure of the hydrogen gas and the vapor pressure of water.

We can use Dalton's law of partial pressures to calculate the partial pressure of the hydrogen gas, and then use the ideal gas law to find the volume of dry hydrogen gas under standard conditions.

First, let's calculate the partial pressure of the hydrogen gas:

The total pressure is 757 torr, and the vapor pressure of water is 19 torr. So the partial pressure of the hydrogen gas is 757 torr - 19 torr = 738 torr.

Next, we can use the ideal gas law to find the volume of dry hydrogen gas under standard conditions:

P1V1/T1 = P2V2/T2
V2 = V1 * (P1/P2) * (T2/T1) = 564 mL * (738 torr/760 torr) * (273.15 K/294.15 K)

V2 ≈ 548 mL

Therefore, the volume of the dry hydrogen gas under standard conditions is approximately 548 mL.