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A saturated hydrocarbon having molecular formula CnH2n+2 diffuses through a porous membrane twice as fast as sulphur dioxide. Calculate the volume occupied by the hydrocarbon at27°C and 2atm ?
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pstnonsonjoku

Answer:

12.3 L

Explanation:

Now we have that the rate of diffusion of the saturated hydrocarbon is R1

Rate of diffusion of sulphur dioxide is R2

Molar mass of hydrocarbon is M1

Molar mass of sulphur dioxide is 64 gmol-1

From Graham's law;

R1/R2 = √64/M1

2/1 =√64/M1

(2/1)^2 = (√64/M1)^2

4/1 = 64/M1

4M1 =64

M1 = 16

To obtain the number of moles of the gas;

(n*12) + (2n + 2) 1 = 16

12n + 2n + 2 = 16

14n + 2 = 16

14n = 16 - 2

n = 14/14

n = 1

Hence the hydrocarbon is CH4

Volume occupied by CH4 at STP = 22.4 L

Hence;

P1 = 1 atm

T1 = 273 K

V1 = 22.4 L

T2 = 300 K

P2 = 2 atm

V2 = ?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 1 * 22.4 * 300/2 * 273

V2 = 12.3 L

Eduard22sly

The volume occupied by the hydrocarbon at 27 °C and 2 atm is 12.31 L

Let the hydrocarbon be initially at standard temperature and pressure (STP)

Thus, we can obtain the new volume of the hydrocarbon by using the combined gas equation as illustrated below:

  • Initial volume (V₁) = STP = 22.4 L
  • Initial pressure (P₁) = STP = 1 atm
  • Initial temperature (T₁) = STP = 273 K
  • Final temperature (T₂) = 27 °C = 27 + 273 = 300 K
  • Final pressure (P₂) = 2 atm
  • Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

(1 × 22.4) / 273 = (2 × V₂) / 300

22.4 / 273 = (2 × V₂) / 300

Cross multiply

273 × 2 × V₂ = 22.4 × 300

546 × V₂ = 6720

Divide both side by 546

V₂ = 6720 / 546

V₂ = 12.31 L

Thus, the volume occupied by the hydrocarbon at 27 °C and 2 atm is 12.31 L

Learn more about gas laws:

https://brainly.com/question/25967416

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