Question

6. While cleaning out a laboratory, a chemist found an unlabeled gas cylinder containing 7.5 x 10 mL of gas at 20.5°C at 1.50 atm of pressure. The chemist believed the mystery gas could be pure He, Ne, Ar, or Kr. Towards this end, the gas was found to have a mass of 1.88 g. What is the molar mass? What is the most likely identity of the mystery gas? 7. Assuming all gas is removed from the tank when filling balloons, how many 0.75 L balloons can be filled from a tank that containing 375 g CO, at 1.0 atm of pressure at a temperature of 298 K?

6 and 7 please explain

Answer 1

Answer 6 -

Given,

Volume of gas = 7.5 * 102 mL or 0.75 L   [1 mL = 10-3 L]

Temperature = 20.5 $\degree$ C or 293.65 K [°C + 273.15 = K]

Pressure = 1.50 atm

Mass = 1.88 g

Molar Mass = ?

Identity of gas = ?

We know that,

PV = (mRT) / MM

where,

P = Pressure

V = Volume

m = mass

R = Gas Constant (0.08206 L⋅atm⋅K−1⋅mol−1)

T = Temperature

MM = Molar Mass

1.50 atm * 0.75 L = (1.88 g*0.08206 L⋅atm⋅K−1⋅mol−1*293.65 K) / MM

1.125 atm.L = (45.30 gL⋅atm⋅mol−1​​​​​​​) / MM

MM = (45.30 gL⋅atm⋅mol−1​​​​​​​) / 1.125 atm.L

MM = 40.3 g / mol [ANSWER]

Argon have molar mass of 39.95 g/mol

So, This gas is Argon. [ANSWER]

Answer 7 -

Given,

Volume of balloons = 0.75 L

Mass of CO2 = 375 g

Pressure = 1.0 atm

Temperature = 298 K

Number of Baloons = ?

We know that,

PV = (mRT) / MM

where,

P = Pressure

V = Volume

m = mass

R = Gas Constant (0.08206 L⋅atm⋅K−1⋅mol−1)

T = Temperature

MM = Molar Mass

PV = (mRT) / MM

1.0 atm * V = (375 g * 0.08206 L⋅atm⋅K−1⋅mol−1 * 298K ) / (44.01 g/mol)

V = 208.37 L

Now,

Number of Balloons = Total Volume / Volume of balloon

Number of Balloons = 208.37 L/ 0.75 L

Number of Balloons = 278 (approx) [ANSWER]