Equilibrium is established between Br2(l) and Br2(g) at 25.0*C. A 250.0 mL sample of the vapor weights 0.486g. What is the vapor pressure of bromine at 25.0*C, in mmHg?
mols Br2 gas = grams/molar mass = ?
Then PV = nRT and solve for P.
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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 = number of moles
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 (in Kelvin) = 25.0°C + 273.15 = 298.15 K
Next, we need to calculate the number of moles of bromine using its molar mass. The molar mass of bromine (Br2) is 159.808 g/mol.
Number of moles (n) = mass (in grams) / molar mass (in grams/mole)
n = 0.486 g / 159.808 g/mol = 0.00304 mol (rounded to 4 decimal places)
Now we have all the information required to solve for the vapor pressure:
PV = nRT
P * 0.250 L = (0.00304 mol) * (0.0821 L·atm/mol·K) * 298.15 K
Simplifying:
P * 0.25 L = 0.07376 L·atm
P = 0.07376 L·atm / 0.25 L
P ≈ 0.295 mmHg (rounded to 3 decimal places)
Therefore, the vapor pressure of bromine at 25.0°C is approximately 0.295 mmHg.
To find the vapor pressure of bromine at 25.0°C, we need to use the ideal gas law equation:
PV = nRT,
where:
P is the pressure in atmosphere (atm),
V is the volume in liters (L),
n is the number of moles,
R is the ideal gas constant (0.0821 L·atm/(mol·K)), and
T is the temperature in Kelvin (K).
First, we need to convert the volume of the sample from milliliters (mL) to liters (L):
250.0 mL ÷ 1000 mL/L = 0.2500 L.
Next, we calculate the number of moles using the molar mass of bromine (Br2):
Molar mass of Br2 = 2(atomic mass of bromine) = 2(79.90 g/mol) = 159.80 g/mol.
Number of moles (n) = mass ÷ molar mass = 0.486 g ÷ 159.80 g/mol = 0.00304 mol.
Now, we convert the temperature from Celsius (°C) to Kelvin (K):
Temperature in Kelvin (T) = 25.0°C + 273.15 = 298.15 K.
Now, we can substitute the values into the ideal gas law equation:
P × 0.2500 L = 0.00304 mol × 0.0821 L·atm/(mol·K) × 298.15 K.
Simplifying the equation:
P × 0.2500 = 0.00304 × 0.0821 × 298.15,
P × 0.2500 = 0.6037.
Finally, solving for P:
P = 0.6037 ÷ 0.2500 ≈ 2.41 atm.
However, the question asks for the vapor pressure of bromine in mmHg. We can convert atmospheres (atm) to millimeters of mercury (mmHg) using the conversion factor:
1 atm = 760 mmHg.
So, the vapor pressure of bromine at 25.0°C is:
P = 2.41 atm × 760 mmHg/atm ≈ 1830 mmHg.
Therefore, the vapor pressure of bromine at 25.0°C is approximately 1830 mmHg.