What is the molecular formula for a compound with the empirical formula: CaCl2 and a molecular mass of 330g.

The empirical formula is CaCl2, which means Ca= x & Cl= 2x & where the molecular weight is 330 g

So x(Ca) + 2x(Cl) = 330 and when the molecular weight of Ca=40 & the molecular weight of Cl = 35

∴ x(40) + 2x (35) =330

  40 x + 70 x = 330

  110 x = 330

   x= 330/110 = 3

∴ the molecular formula is Ca3Cl6

Explanation:

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%