Consider the unbalanced chemical equation: H2SO4 (aq) + Fe(OH)3 (S) → Fe2(SO4)3 (S) + H2O (L)
A volume of 38.0 mL of aqueous H2SO4 solution was required to
react completely with 0.685 g Fe(OH)3 (molar mass = 106.8 g/mol)
to produce Fe2(SO4)3. Calculate the molar concentration of the
H2SO4 solution.
A) 0.253 M
B) 0.344 M
C) 0.214 M
D) 0.301 M
E) 0.175 M

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%

yes they ca   is ur ansswerrr

3

explanation: