QJEEM20Ch5eff: JEE Main 2020 Chemistry - Class 11 JEEM20Ch5eff Solution: 6.023 × 10 22 molecules are present in 10 g of a substance 'x'. …

Mass of 6.023 × 10²² molecules of a substance = 10 g

Mass of 6.023 × 10²³ molecules of the substance = 100 g

$$\therefore$$ Molar mass of the substance = 100 g mol^–1

Molarity of the solution =

moles of solute

volume of solution(in l)

= $$\frac{\left(5/100\right)}{2}$$ = 0.025