What volume does 4.24g of argon gas occupy at 58.2 degrees celcuis and 1528torr

The volume occupied by the argon gas is, 1.44 L

Explanation :

To calculate the volume of argon gas we are using ideal gas equation:

where,

P = pressure of argon gas = 1528 torr = 2.010 atm

Conversion used : (1 atm = 760 torr)

V = volume of argon gas = ?

T = temperature of argon gas =

R = gas constant = 0.0821 L.atm/mole.K

w = mass of argon gas = 4.24 g

M = molar mass of argon gas = 39.95 g/mole

Now put all the given values in the ideal gas equation, we get:

Therefore, the volume of occupied by argon gas is 1.44 L

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%