Question

6*10^20 molecules of urea is present in 100ml of its solution.The conc of urea is

Answer 1

Moles of urea = molecules/avogadro no

                       =    6.02X10^20 / 6.022 X 10^23

                       =          10^-3 moles

So it will have 10^-2 mole in 1 litre of solution

because 

Molarity/concentration = mole / litre

                                         = 10^-3/1L

                                        =    10^-2M