1. When gases are collected using displacement of water, some water vapor mixes in with the gas being collected. A student performed a similar experiment in which 35.2 ml of H2 gas was collected over water at a temperature of 21 degrees Celsius. If the barometric pressure was 758mmHg. How many grams of water vapor mixed in with the hydrogen gas.
bv b
To determine the number of grams of water vapor that mixed in with the hydrogen gas, we need to use the ideal gas law equation. The ideal gas law equation is:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L atm / mol K)
T = temperature of the gas (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin:
T = 21°C + 273.15 = 294.15 K
Next, we need to convert the given pressure from mmHg to atm:
P = 758 mmHg / 760 mmHg/atm = 0.9974 atm
Now, we have the pressure (P), temperature (T), and volume (V) of the hydrogen gas. However, we need to consider the presence of water vapor in the collected gas. Since water vapor is mixed with the hydrogen gas, the total pressure (P_total) is the sum of the pressure of hydrogen gas (P_hydrogen) and the vapor pressure of water at the given temperature.
To find the vapor pressure of water at 21°C, we can refer to a vapor pressure table (e.g., the Antoine equation). Let's assume the vapor pressure of water at 21°C is 18.6 mmHg.
Now, we can calculate the partial pressure of hydrogen gas (P_hydrogen) using the following equation:
P_hydrogen = P_total - P_water_vapor
P_hydrogen = 0.9974 atm - (18.6 mmHg / 760 mmHg/atm) = 0.9974 - 0.0245 atm = 0.9729 atm
Since we have the pressure and volume of hydrogen gas, we can calculate the number of moles of hydrogen gas (n) using the ideal gas law by rearranging the equation:
n = PV / RT
n = (0.9729 atm * 0.0352 L) / (0.0821 L atm / mol K * 294.15 K) ≈ 0.00152 moles of hydrogen gas
Now, we need to find the number of moles of water vapor that mixed in with the hydrogen gas. Since water vapor is also a gas and behaves according to the ideal gas law, we can assume that the water vapor is at the same temperature and pressure as the total collected gas. Therefore, the number of moles of water vapor (n_water_vapor) is equal to the number of moles of the total collected gas (n_total_gas) minus the number of moles of hydrogen gas (n_hydrogen):
n_water_vapor = n_total_gas - n_hydrogen
We already have n_hydrogen (0.00152 moles of hydrogen gas). To find n_total_gas, we can use the ideal gas law again, this time with the given total volume (V_total):
n_total_gas = PV_total / RT
n_total_gas = (0.9974 atm * 0.0352 L) / (0.0821 L atm / mol K * 294.15 K) ≈ 0.00154 moles of total collected gas
Now, we can find n_water_vapor:
n_water_vapor = 0.00154 moles - 0.00152 moles ≈ 0.00002 moles of water vapor
To determine the mass of water vapor mixed in with the hydrogen gas, we need to use the molar mass of water (H2O), which is approximately 18.015 g/mol. The mass (m_water_vapor) is calculated by multiplying the number of moles (n_water_vapor) by the molar mass (M_water):
m_water_vapor = n_water_vapor * M_water
m_water_vapor = 0.00002 moles * 18.015 g/mol ≈ 0.00036 g of water vapor
Therefore, approximately 0.00036 grams of water vapor mixed in with the hydrogen gas in this experiment.