A 125 mL flask contains 75 kPa Cl2 at 22 ˚C.
(a) How many moles of Cl2 are in the flask?
(b) What mass of Cl2 is in the flask?
To solve this problem, we can use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
(a) To find the number of moles of Cl2 in the flask, we need to rearrange the Ideal Gas Law equation to solve for n:
n = PV / RT
First, we need to convert the given volume from milliliters (mL) to liters (L) and the given pressure from kilopascals (kPa) to atmospheres (atm).
1 L = 1000 mL
1 atm = 101.3 kPa
125 mL / 1000 mL/L = 0.125 L
75 kPa * (1 atm / 101.3 kPa) = 0.740 atm
Next, we need to convert the temperature from degrees Celsius (˚C) to Kelvin (K).
T(K) = T(˚C) + 273.15
T(K) = 22 ˚C + 273.15 = 295.15 K
Now we can substitute the values into the equation:
n = (0.740 atm) * (0.125 L) / [(0.0821 Latm/molK) * (295.15 K)]
n ≈ 0.0031 moles
Therefore, there are approximately 0.0031 moles of Cl2 in the flask.
(b) To find the mass of Cl2 in the flask, we can use the molar mass of Cl2, which is 70.90 g/mol.
m = n * M
m ≈ 0.0031 moles * 70.90 g/mol
m ≈ 0.218 g
Therefore, there are approximately 0.218 grams of Cl2 in the flask.
To answer these questions, we can use the ideal gas law equation, which relates the pressure, volume, temperature, and number of moles of a gas.
The ideal gas law equation is given by:
PV = nRT
Where:
P = pressure of the gas (in Pa)
V = volume of the gas (in m^3)
n = number of moles of the gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature of the gas (in K)
Let's solve each part of the question step by step.
(a) How many moles of Cl2 are in the flask?
First, we need to convert the volume of the flask from mL to m^3 since the units in the ideal gas law equation need to be consistent.
1 mL = 1 x 10^-6 m^3
So, the volume of the flask in m^3 is:
V = 125 mL * (1 x 10^-6 m^3/1 mL) = 0.000125 m^3
Next, we convert the temperature from ˚C to Kelvin (K).
T(K) = T(˚C) + 273.15
T(K) = 22 ˚C + 273.15 = 295.15 K
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
75 kPa * 1000 = n * 8.314 J/(mol·K) * 0.000125 m^3 * 295.15 K
Simplifying,
n = (75 kPa * 1000) / (8.314 J/(mol·K) * 0.000125 m^3 * 295.15 K)
n ≈ 3.00 moles
Therefore, there are approximately 3.00 moles of Cl2 in the flask.
(b) What mass of Cl2 is in the flask?
To determine the mass of Cl2, we can use the molar mass of Cl2, which is the sum of the atomic masses of chlorine (Cl). The molar mass of Cl2 is approximately 70.9 g/mol.
m = n * M
where:
m = mass of Cl2 (in grams)
n = number of moles of Cl2
M = molar mass of Cl2
m = 3.00 moles * 70.9 g/mol
m ≈ 212.7 grams
Therefore, there is approximately 212.7 grams of Cl2 in the flask.
1 mole at those conditions would occupy 32.7L
So, you must have 0.125/32.7 = 0.0038 moles
now multiply that by 35.5 g/mole