Question

A 10.00L sample of gas has a mass of 11.92g at stp. What is the molar mass of the gas?

Answer 1

• As we know STP means Standard Temperature and Pressure. Where, Pressure is 1 atm and temperature is 273 K and also Ideal gas law ( PV=nRT) or (PV = w/M RT)

Where –

• P = Pressure in atm
• V = Volume in L
• n = moles
• R = Ideal gas law constant
• T = Temperature in K
• w = Given Mass
• M = Molar Mass

Now, according to the question –

• w =11.92 g
• V = 10 L
• P = 1 atm
• T = 273 K
• R = 0.0821 atm L/ mol K

Calculation –

`\qquad`
`\pink{\twoheadrightarrow\bf PV = nRT}`

`\qquad`
`\pink{\twoheadrightarrow\bf PV = (w)/(M)* RT}`

`\qquad`
`\twoheadrightarrow\sf(w)/(M) = (PV)/(RT)`

`\qquad`
`\twoheadrightarrow\sf (M)/(w) = (RT)/(PV)`

`\qquad`
`\twoheadrightarrow\sf M = (RT)/(PV)* w`

`\qquad`
`\twoheadrightarrow\sf M = (0.0821 * 273 )/(1 * 10 ) * 11.92`

`\qquad`
`\twoheadrightarrow\sf M = (22.4)/(10) * 11.92`

`\qquad`
`\twoheadrightarrow\sf M = 2.24* 11.92`

`\qquad`
`\twoheadrightarrow\sf M = 26.7`

`\qquad`
`\pink{\twoheadrightarrow\bf M = 26.7\: g}`

• Henceforth, Molar Mass is 26.7 g

_______________________________________

Answer 2

Let's assume that the gas is an ideal gas.

1 mole of ideal gas has the volume as 22.4 L at STP (273 K , 1 atm)

The given gas has 10.00 L. Hence we can find out the moles of gas.

moles in 22.4 L gas = 1 mol
moles in 10.00 L gas =
`(1 mol * 10.00 L)/(22.4 L)` = 0.446 mol

moles = mass / molar mass

hence, molar mass = mass / moles
= 11.92 g / 0.446
= 26.73 g/mol

Hence, the molar mass of the gas is 26.73 g/mol