A volatile liquid sample is placed in an Erlenmeyer flask that has been sealed except for a tiny hole that allows vapor to escape. Upon heating, the liquid is converted to vapor that fills the flask. Cooling the flask allows the vapor to condense back to liquid. The mass of the condensed liquid is found to be 0.608g. The actual volume of the flask is 158.5 mL. The temperature of the flask contents is 32.4 degrees C and the atmospheric pressure reading is 30.52 inches of mercury. From this data determine the molecular weight of the volatile liquid.
PV = nRT. You have P (in inches which must be converted to atmospheres), you have V (convert to L), R and T(convert to kelvin). Solve for n = mols.
Then mols = grams/molar mass. You solve for molar mass.
To determine the molecular weight of the volatile liquid, we can use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, we need to convert the temperature from degrees Celsius to Kelvin. The equation is:
T(K) = T(°C) + 273.15
T(K) = 32.4 + 273.15 = 305.55 K
Now, we need to convert the atmospheric pressure from inches of mercury to atm. The conversion factor is:
1 atm = 29.92 inches of mercury
P = 30.52 inches of mercury * (1 atm / 29.92 inches of mercury) = 1.019 atm
Next, we need to calculate the number of moles of the volatile liquid. Rearranging the ideal gas law equation, we can solve for n:
n = PV / RT
n = (P * V) / (R * T)
n = (1.019 atm * 0.1585 L) / (0.0821 L·atm/(mol·K) * 305.55 K) = 0.00678 mol
Now, we can calculate the molecular weight (molar mass) of the volatile liquid by dividing the mass of the condensed liquid by the number of moles:
Molecular weight = Mass / Moles
Molecular weight = 0.608 g / 0.00678 mol = 89.71 g/mol
Therefore, the molecular weight of the volatile liquid is approximately 89.71 g/mol.