the ratio of
partial pressure of CH4 and O2 at 500 torr and 100 torr pressure in 10dm3 vessel is/
?

2 thoughts on “the ratio of<br /> partial pressure of CH4 and O2 at 500 torr and 100 torr pressure in 10dm3 vessel is/<br /> ?”

1. igyhloydhtsditdfppft a few f

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2. Answer:

   Firstly to find the pressure, begin by finding the Mr of methane and oxygen

   Mr is the relative molecular mass

   Mr of methane:

   The formula is – CH4

   Mr = 12×1 + 1×4

   = 12 + 4

   = 16

   Mr of oxygen:

   The formula is O2 ( 2 is in the subscript)

   Mr = 16×2

   = 32

   So the ratio of the masses of methane : oxygen

   = 16 : 32

   = 1 : 2

   To calculate the force, we will need to multiply the mass with the gravitational field strength that is approximately 9.8N/kg but here both the molecules will experience the same field strength so we can ignore it.

   So the ratio of mass is

   CH4 : O2

   = 1 : 2

   Now as given in the question that the number of molecules are in the ratio

   methane : oxygen

   = 2 : 1

   Then multiply the mass and number of molecules ratio

   For methane:

   = mass × molecules

   = 1 × 2

   = 2

   For oxygen:

   = mass × molecules

   = 2 × 1

   = 2

   Thus the pressure is in the ratio =

   methane : oxygen

   = 2 : 2

   = 1 : 1

   So as an extension to this thought, the pressure exerted by oxygen and methane will be equal.

   Therefore, oxygen exerts 1÷2 or 50% pressure

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