Tom has 1 cup of a 5% vinegar-water solution, 2 cups of a 10% vinegar-water solution, and 3 cups of a 15% vinegar-water solution. He needs at least 8 cups of a 5% vinegar-water solution for his chemistry project. What is the number of cups of pure water he should add to the total of all his vinegar-water solutions to complete the project?

Another eight cups of pure water will be required.

Explanation:

Start by calculating the amount (in cups) of vinegar in these six cups:

Therefore, if these six cups of solutions were mixed, the mixture would contain cups of vinegar. Also, the mixture is supposed to have a volume of six cups. Therefore, the concentration of this six-cup mixture would be:

.

In other words, if these six cups of vinegar-water solutions are mixed, it would be necessary to dilute (by adding more water) before reaching the concentration of .

The goal is to dilute this solution with cups of vinegar to a concentration of . Consider: what would be the volume the solution after dilution?

The volume of a solution can be found from the quantity and concentration of its solute:

After dilution, the volume of this solution would be:

.

At this moment, this solution only has a volume of six cups. Therefore, cups of pure water will be required.

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%