Question

6.025*108 molecules of acetic acid present in 500 ml in its solution. the concentration of solution is

Answer 1

6.025*10^8 molecules of acetic acid present in 500 ml of its solution

One mole of a substance contains the Avogadro's number of particles

Avogadro's number = 6.022 x 10^23


Number of moles of acetic acid = (6.025 x 10^8)/(6.022 x 10^23)

                                                  = 1 x 10^-15

Volume = 500/1000 = 0.5 litres


Molarity = Number of moles/Volume in litres

             = (1 x 10^-15) / 0.5

             = 2 x 10^-15 M

Answer 2

Answer:

Explanation:

6.025*10^8 molecules of acetic acid present in 500 ml of its solution

One mole of a substance contains the Avogadro's number of particles

Avogadro's number = 6.022 x 10^23

Number of moles of acetic acid = (6.025 x 10^8)/(6.022 x 10^23)

= 1 x 10^-15

Volume = 500/1000 = 0.5 litres

Molarity = Number of moles/Volume in litres

= (1 x 10^-15) / 0.5

= 2 x 10^-15 M