What is the molarity of a stock solution if 60 mL were used to make 150 mL of a .5M solution

The answer to your question is 1.25 M

Explanation:

Data

Molarity 1 = ?

Volume 1 = 60 ml

Molarity 2 = 0.5 M

Volume 2 = 150 ml

Process

1.- Write the dilution formula

       Molarity 1 x Volume 1 = Molarity 2 x Volume 2

-Solve for Molarity 1

       Molarity 1 = Molarity 2 x Volume 2 / Volume 1

-Substitution

        Molarity 1 = (0.5)(150) / 60

-Simplification

        Molarity = 75 / 60

-Result

        Molarity = 1.25 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%

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.