Question

name Prelaboratory Assignment: Gas Laws Technical objective: A flask with a sample of gas at room temperature and pressure weighs 97.4842 g. The same flask with air (density 1.17 g/L) weighs 96.8848 g. The same flask filled with water ( 0.998 g/mL) weighs 345.86 g. The pressure in the room is 0.989 atm and the temperature is 22.2°C. 1. Calculate the volume of the flask. 2. Calculate the mass of the air that the flask can hold. 3. Calculate the mass of the empty (no air) flask. 4. Calculate the mass of the gas in the flask. 5. Use the ideal gas law to calculate the moles of gas in the flask. 6. Calculate the molar mass of the gas in the flask. 109

Answer 1

1)

Let the mass of the flask be M

Let V be the volume of Flask

Density of air = 1.17 g/L (given)

Density of water = 998 g/L (given)

Total mass of Flask + water = 345.86 g

M + density of water * volume of flask = 345.86

M + 998*V = 345.86 -------(1)

Total mass of Flask + air = 96.8848 g

M + 1.17*V = 96.8848 --------(2)

subtracting equation (2) from equation (1) we get

M + 998*V - M - 1.17*V = 345.86 - 96.8848

(998-1.17) * V = 248.9752

996.83*V = 248.9752

=> V = 248.9752/996.83

=> Volume of Flask, V = 0.249767 L

2) Mass of air = density of air * Volume of flask

=> Mass of air = 1.17 * 0.249767

=> mass of air = 0.292227 g

3) Mass of empty flask = Mass of flask + air - Mass of air = 96.8848-0.292227

Mass of empty flask = 96.59257 g

4) Mass of gas in flask = Mass of Flask + Gas - Mass of flask = 97.4842 - 96.59257

Mass of gas in flask = 0.891627 g

5) PV = nRT

=> n = PV/RT

=> n = 0.989*0.249767/(0.0821*295.35)

=> n = 0.010187 moles

6) Molar mass of gas = mass of gas/number of moles of gas = 0.891627/0.010187

=> molar mass of gas = 87.52501 g/mole