Respuesta :

Ishankahps
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 = [tex] \frac{1 mol * 10.00 L}{22.4 L} [/tex] = 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
  • 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 –

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf PV = nRT}[/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf PV = \dfrac{w}{M}\times RT}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf\dfrac{w}{M} = \dfrac{PV}{RT}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{M}{w} = \dfrac{RT}{PV}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf M = \dfrac{RT}{PV}\times w[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf M = \dfrac{0.0821 \times 273 }{1 \times 10 } \times 11.92 [/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf M = \dfrac{22.4}{10} \times 11.92[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf M = 2.24\times 11.92[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf M = 26.7 [/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf M = 26.7\: g}[/tex]

  • Henceforth, Molar Mass is 26.7 g

_______________________________________

Otras preguntas