g If 0.600 g of a gas occupies 211 mL at 24 oC and 660 mmHg of pressure, what is the molar mass of the gas

79.89 g/mol

Explanation:

The following data were obtained from the question:

Mass (m) of gas = 0.6 g

Volume (V) = 211 mL

Temperature (T) = 24 °C

Pressure (P) = 660 mmHg

Molar mass of gas =.?

Next, we convert 24 °C to Kelvin temperature. This can be obtained as follow:

Temperature (K) = temperature (°C) + 273

Temperature (°C ) = 24 °C

Temperature (K) = 24 + 273 = 297 K

Next, we shall convert 211 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

211 mL = 211 mL × 1 L / 1000 mL

211 mL = 0.211 L

Next, we shall convert 660 mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

660 mmHg = 660 mmHg × 1 atm / 760 mmHg

660 mmHg = 0.868 atm

Next, we shall determine the number of mole of the gas. This can be obtained as shown below:

Volume (V) = 0.211 L

Temperature (T) = 297 K

Pressure (P) = 0.868 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.868 × 0.211 = n × 0.0821 × 297

Divide both side by 0.0821 × 297

n = (0.868 × 0.211) / (0.0821 × 297)

n = 7.51×10¯³ mole.

Finally, we shall determine the molar mass of the gas as shown below:

Mass of gas = 0.6 g

Number of mole of gas = 7.51×10¯³ mole.

Molar mass of gas =.?

Mole = mass /Molar mass

7.51×10¯³ = 0.6 / Molar mass

Cross multiply

7.51×10¯³ × molar mass = 0.6

Divide both side by 7.51×10¯³

Molar mass = 0.6 / 7.51×10¯³

Molar mass of gas = 79.89 g/mol

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.