A sample of a volatile liquid is vaporized completely in a 240 ml falsh at 99 c and 757 torr. The vapor is then allowed to condense to al liquid and its mass is determined to be .564 grams.
What is the molecular mass of the liquid (that was vaporized)?
Use PV=nRT to find moles
Then, mole mass=mass/moles
To find the molecular mass of the liquid, we need to use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. In this case, we can assume that the vapor behaves ideally.
First, let's determine the number of moles of the vapor using the ideal gas law equation:
PV = nRT
Where:
P = pressure (757 torr)
V = volume (240 mL or 0.24 L)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L⋅atm/mol⋅K)
T = temperature in Kelvin (99 °C + 273.15 °C = 372.15 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
Substituting the given values, we get:
n = (757 torr * 0.24 L) / (0.0821 L⋅atm/mol⋅K * 372.15 K)
n ≈ 0.0653 mol
Now, we can calculate the molecular mass using the equation:
Molecular mass = mass / moles
Substituting the given mass (0.564 grams) and number of moles (0.0653 mol), we get:
Molecular mass = 0.564 g / 0.0653 mol
Molecular mass ≈ 8.64 g/mol
Therefore, the molecular mass of the liquid (that was vaporized) is approximately 8.64 g/mol.