You have 229 mL of chlorine trifluoride gas at 735 mmHg and
30
°
C. What is the mass (in g) of the sample?
g
To determine the mass of the sample, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure in atm (convert mmHg to atm by dividing by 760: 735 mmHg / 760 = 0.966 atm)
V = volume in liters (convert mL to L by dividing by 1000: 229 mL / 1000 = 0.229 L)
n = number of moles of gas
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin (convert Celsius to Kelvin: 30 + 273 = 303 K)
Rearranging the equation to solve for n:
n = (PV)/(RT)
n = (0.966 atm * 0.229 L) / (0.0821 L·atm/(mol·K) * 303 K)
n = 0.3004 mol
Since the molar mass of chlorine trifluoride (ClF3) is approximately 92.45 g/mol, we can calculate the mass of the sample:
mass = n * molar mass
mass = 0.3004 mol * 92.45 g/mol
mass ≈ 27.64 g
Therefore, the mass of the sample is approximately 27.64 g.
To determine the mass of the sample, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given volume from milliliters to liters:
229 mL = 229/1000 L
V = 0.229 L
Next, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 30 + 273.15
T = 303.15 K
Now we can rearrange the ideal gas law equation to solve for n:
n = PV / (RT)
Substituting the given values:
P = 735 mmHg = 735/760 atm (since 1 atm = 760 mmHg)
R = 0.0821 L·atm/(mol·K)
n = (735/760) * 0.229 / (0.0821 * 303.15)
n ≈ 0.0103 mol
Finally, to calculate the mass (in grams) of the sample, we need to use the molar mass of chlorine trifluoride (ClF3), which is approximately 92.45 g/mol.
Mass = n * molar mass
Mass = 0.0103 mol * 92.45 g/mol
Mass ≈ 0.951 g
Therefore, the mass of the sample is approximately 0.951 grams.