A student weighs an empty flask and stopper and finds the mass to be 53.256 g. She then adds about 5 mL of an unknown liquid and heats the flask in a boiling water bath at 98.8 degree C. After all the liquid is vaporized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momentarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of 53.780 g. The volume of the flask is known to be 231.1 mL. The absolute atmospheric pressure in the laboratory that day is 728 mm Hg.

a. What was the pressure of the vapor in the flask in atm?
b. What was the temperature of the vapor in K? The volume of the flask in liters?
c. What was the mass of condensed vapor that was present in the flask?
d. How many moles of condensed vapor were present?
e. What is the mass of one mole of vapor?

a) 0.957895 atm

b) 372.05 K

c) 0.524 grams

d)  0.00725 moles

e) Molar mass gas = 30.54 g/mol

Explanation:

Step 1: Data given

Mass of the flask = 53.256 grams

Volume = 5 mL

Temperature of water = 98.9 °C = 372.05 K

Mass obtained = 53.78 grams

The volume of the flask is known to be 231.1 mL = 0.2311 L

The absolute atmospheric pressure in the laboratory that day is 728 mm Hg.

Step 2: Calculate the mass of the gas

Mass gas = 53.780 - 53.256 = 0.524 grams

Step 3: Calculate moles

p*V = n*R*T

⇒with p = the pressure = 0.957895 atm

⇒with V = the volume = 0.2311 L

⇒with n = the number of moles gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol * K

⇒with T = the temperature = 372.05 K

n = (p*V)/(R*T)

n= (0.957895  * 0.2311) / (0.08206 * 372.05)

n = 0.00725 moles

Step 4: Calculate mass of the gas

Molar mass = mass / moles

Molar mass gas = 0.524 grams . 0.00725 moles

Molar mass gas = 30.54 g/mol

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%