Question

A student poured a 1.5 mL sample of volatile liquid into a 250 ml flask whose mass was 90.008 grams. The flask was sealed with a piece of foil, immersed in a boiling water bath, and heated for several minutes. The liquid vaporized, and it warmed to 99.0 C, the temperature of the boiling water. Air and some unknown vapor escaped from the flask so that the pressure of the vapor was equal to the barometric pressure of the lab, 756 mmHg. The flask was removed from the boiling water bath and cooled. The vapor condensed, and the mass of the flask with the condensed liquid was 92.209. The student emptied the flask, filled it with water, and determined the mass of the flask with water was 360.89g.

1) Pressure of the vapor in atm
2)Temperature of the vapor in Kelvin
3) Volume of the vapor (assume the density of h2o= 1.00g/ml)
4) Mass of the condensed vapor
5) Moles of vapor
6) Molar mass of volatile liquid

Answer 1

Answer: Therefore, the molar mass of the volatile liquid is 176.08 g/mol.

Explanation:

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 gas constant, and T is the temperature in Kelvin.

Pressure of the vapor in atm:

The pressure of the vapor is given as 756 mmHg. Converting this to atm:

1 atm = 760 mmHg

756 mmHg = 0.9947 atm

Therefore, the pressure of the vapor is 0.9947 atm.

Temperature of the vapor in Kelvin:

The temperature of the vapor is given as 99.0 C. Converting this to Kelvin:

T(K) = T(C) + 273.15

T(K) = 99.0 + 273.15

T(K) = 372.15 K

Therefore, the temperature of the vapor is 372.15 K.

Volume of the vapor (assume the density of h2o= 1.00g/ml):

To find the volume of the vapor, we can use the volume of the flask and the mass of the water that was added to it. Since the mass of the flask with the condensed liquid is given as 92.209 g, and the mass of the empty flask is 90.008 g, the mass of the condensed vapor is:

92.209 g - 90.008 g = 2.201 g

Since the density of water is 1.00 g/mL, the volume of the water that was added to the flask is:

2.201 mL

Therefore, the volume of the vapor is 250 mL - 1.5 mL - 2.201 mL = 246.299 mL.

Mass of the condensed vapor:

As calculated earlier, the mass of the condensed vapor is 2.201 g.

Moles of vapor:

To find the number of moles of the vapor, we can use the ideal gas law equation, and solve for n:

n = PV/RT

Since we know the pressure, volume, and temperature of the vapor, we can substitute those values along with the gas constant R:

n = (0.9947 atm)(246.299 mL/1000 mL/L) / (0.08206 Latm/molK)(372.15 K)

Simplifying, we get:

n = 0.0125 mol

Therefore, the number of moles of vapor is 0.0125 mol.

Molar mass of volatile liquid:

To find the molar mass of the volatile liquid, we can use the formula:

molar mass = mass/ moles

The mass of the volatile liquid is the difference between the mass of the flask with the condensed liquid and the mass of the empty flask:

mass = 92.209 g - 90.008 g = 2.201 g

Substituting the values for mass and moles, we get:

molar mass = 2.201 g/ 0.0125 mol

Simplifying, we get:

molar mass = 176.08 g/mol