What is the molecular weight of a gas if a 21.0g sample has a pressure of 836mm Hg at 25.0 degrees C in a 2.00L flash?

234amu

To determine the molecular weight of a gas, you can use the Ideal Gas Law equation:

PV = nRT

Where:
P is the pressure of the gas (in atm)
V is the volume of the gas (in liters)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas (in Kelvin)

First, we need to convert the given pressure from mm Hg to atm.
1 atm = 760 mm Hg

So, 836 mm Hg can be converted to atm as follows:

836 mm Hg × (1 atm / 760 mm Hg) = 1.099 atm (rounded to three decimal places)

Next, we need to convert the temperature from degrees Celsius to Kelvin.
The temperature in Kelvin can be obtained using the equation:

T(K) = T(°C) + 273.15

So, 25.0 degrees Celsius can be converted to Kelvin as follows:

25.0 + 273.15 = 298.15 K (rounded to five decimal places)

Now that we have the pressure (1.099 atm), volume (2.00 L), and temperature (298.15 K), we can rearrange the Ideal Gas Law equation to solve for the number of moles (n) of the gas:

n = PV / RT

Substituting the values:

n = (1.099 atm) × (2.00 L) / [(0.0821 L·atm/(mol·K)) × (298.15 K)]

Simplifying the equation:

n ≈ 0.091 moles (rounded to three decimal places)

Finally, to find the molecular weight, we divide the mass (21.0 g) of the sample by the number of moles (0.091 moles):

molecular weight = mass / moles
molecular weight = 21.0 g / 0.091 moles

molecular weight ≈ 230.77 g/mol (rounded to two decimal places)

Therefore, the molecular weight of the gas is approximately 230.77 g/mol.

To find the molecular weight of a gas, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = Ideal gas constant
T = Temperature of the gas in Kelvin

First, let's convert the given values to the appropriate units:

Pressure:
The given pressure is 836 mmHg. To use this value in the equation, we need to convert it to atmospheres (atm). Since 1 atm = 760 mmHg, we can convert the pressure as follows:

836 mmHg × (1 atm / 760 mmHg) = 1.10 atm

Volume:
The given volume is 2.00 L, which is already in the appropriate unit.

Temperature:
The given temperature is 25.0 degrees Celsius. To convert it to Kelvin, we add 273.15 to the Celsius temperature:

25.0 + 273.15 = 298.15 K

Now that we have all the necessary values in appropriate units, we can rearrange the ideal gas law equation to solve for the number of moles (n) of the gas:

n = PV / RT

Substituting the values into the equation:

n = (1.10 atm) × (2.00 L) / [(0.0821 L·atm/(mol·K)) × (298.15 K)]

Performing the calculation:

n ≈ 0.088 mol

The number of moles of the gas is approximately 0.088 mol.

To find the molecular weight (MW) of the gas, we divide the mass (m) of the gas by the number of moles (n):

MW = m / n

The given mass is 21.0 g, so substituting the values:

MW = 21.0 g / 0.088 mol

MW ≈ 238.6 g/mol

The molecular weight of the gas is approximately 238.6 g/mol.