a sample of vapor weighing 0.745 g filled a 250 ml flask when placed in a water bath at 98 degrees celsius. the barometric pressure that day was 753 torr. calculate the volume of the vapor at stp, and the weight of one mol of the volitile liquid
I would use (P1V1/T1) = (P2V2/T2) to convert volume of 250 at the conditions listed to STP. Convert V to liters, divide by 22.4 to convert to mols, then mol = g/molar mass and solve for molar mass.
To calculate the volume of the vapor at STP (Standard Temperature and Pressure) and the weight of one mole of the volatile liquid, we'll need to use the ideal gas law and the concept of molar mass.
First, let's calculate the volume of the vapor at STP:
1. Convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 98 + 273.15 = 371.15 K
2. Convert the pressure from torr to atmospheres (since STP is defined as 1 atmosphere):
P(atm) = P(torr) / 760
P(atm) = 753 / 760 = 0.9911 atm
3. Apply 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 (0.0821 L·atm/mol·K),
and T is the temperature.
Rearrange the equation to solve for the volume at STP:
V(stp) = (nRT) / P(stp)
At STP, the pressure (P(stp)) is 1 atm, and let's assume the number of moles (n) is 1 mole.
V(stp) = (1 mol * 0.0821 L·atm/mol·K * 371.15 K) / 1 atm
V(stp) = 30.254 L
Therefore, the volume of the vapor at STP is 30.254 liters.
Next, let's calculate the weight of one mole of the volatile liquid:
1. Calculate the number of moles of the vapor:
moles = mass / molar mass
Given the mass of the vapor is 0.745 g, let's find the molar mass.
2. Determine the molar mass:
The molar mass can be obtained from the atomic masses of the individual elements in the compound.
Since you haven't mentioned the specific volatile liquid, I'll provide a general example.
Let's assume the volatile liquid is alcohol (ethanol), which has the chemical formula C2H5OH.
The atomic masses are:
C (carbon) - 12.01 g/mol
H (hydrogen) - 1.01 g/mol
O (oxygen) - 16.00 g/mol
Calculate the molar mass:
molar mass = (2 * atomic mass of C) + (6 * atomic mass of H) + atomic mass of O
molar mass = (2 * 12.01) + (6 * 1.01) + 16.00
molar mass = 46.07 g/mol (approximately)
3. Calculate the number of moles:
moles = mass / molar mass
moles = 0.745 g / 46.07 g/mol
moles = 0.01614 mol (approximately)
Therefore, the weight of one mole of the volatile liquid is approximately 46.07 g/mol, and the volume of the vapor at STP is approximately 30.254 liters.