Equilibrium is established between Br2(l) and Br2(g) at 25.0 (degrees)C. A 250.0 ml sample of the vapor weighs 0.486g. What is the vapor pressure of bromine at 25.0 (degrees)C, in millimeters of mercury.
Can you do this using PV = nRT?
You know V, n(or can calculate n), R, and T. Solve for P.
226.1 mm Hg
To find the vapor pressure of bromine at 25.0°C, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25.0 + 273.15 = 298.15 K
Next, we need to calculate the number of moles of bromine gas. We can use the molar mass of bromine (Br2) to convert grams to moles.
Molar mass of Br2 = 2 × (atomic mass of Br) = 2 × 79.90 g/mol = 159.80 g/mol
moles of Br2 = mass of Br2 / molar mass of Br2
moles of Br2 = 0.486 g / 159.80 g/mol
Now, we can calculate the vapor pressure using the ideal gas law equation.
PV = nRT
P × 0.250 L = (0.486 g / 159.80 g/mol) × (0.0821 L·atm/mol·K) × 298.15 K
P × 0.250 L = (0.004877 mol) × (0.0821 L·atm/mol·K) × 298.15 K
P × 0.250 L = 0.010267
P = 0.010267 / 0.250 L
P = 0.04107 atm
To convert this to millimeters of mercury, we use the conversion factor:
1 atm = 760 mmHg
P(mmHg) = P(atm) × 760 mmHg
P(mmHg) = 0.04107 atm × 760 mmHg
P(mmHg) = 31.25 mmHg
Therefore, the vapor pressure of bromine at 25.0°C is approximately 31.25 mmHg.
To find the vapor pressure of bromine at 25.0 degrees Celsius, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (in units of force per unit area)
V is the volume of the gas (in units of cubic meters)
n is the number of moles of the gas
R is the ideal gas constant (8.31 J/(mol·K) or 0.0821 L·atm/(mol·K))
T is the temperature of the gas (in units of Kelvin)
First, we need to determine the number of moles of bromine vapor in the 250.0 ml sample. We can use the equation:
n = m/M
Where:
n is the number of moles
m is the mass of the substance (in units of grams)
M is the molar mass of the substance (in units of grams per mole)
Given that the mass of the sample is 0.486g, and the molar mass of bromine (Br2) is 159.808 g/mol, we can now calculate the number of moles:
n = 0.486 g / 159.808 g/mol ≈ 0.003034 mol
Now, let's convert the volume from milliliters to liters:
V = 250.0 ml * (1 L / 1000 ml) = 0.250 L
Next, we convert the temperature from Celsius to Kelvin by adding 273.15 to the value:
T = 25.0°C + 273.15 ≈ 298.15 K
Finally, we can substitute the values into the ideal gas law equation to solve for P:
P * 0.250 L = 0.003034 mol * 0.0821 L·atm/(mol·K) * 298.15 K
P * 0.250 L = 0.07604682 L·atm
P = (0.07604682 L·atm) / 0.250 L
P ≈ 0.30419 atm
To convert the pressure from atm to millimeters of mercury (mmHg), we can use the conversion factor:
1 atm ≈ 760 mmHg
Therefore, the vapor pressure of bromine at 25.0 degrees Celsius is approximately:
P ≈ 0.30419 atm * 760 mmHg/atm
P ≈ 231.51744 mmHg
So, the vapor pressure of bromine at 25.0 degrees Celsius is approximately 231.5 mmHg.