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Mole Concept Basic - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • MOLE CONCEPT AND MOLAR MASS is considered one the most difficult concept.

  • 32 Questions around this concept.

Solve by difficulty

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

The number of water molecules is maximum in :

If Avogadro number NA, is changed from 6.022 x 1023 mol-1 to 6.022 x 1020 mol-1, this would change :

Concepts Covered - 1



  • A mole is a unit that represents 6.023 x 1023 particles, atoms, molecules or ions, etc., irrespective of their nature.

  • Mole is related to the mass of the substance, the volume of gaseous substance and the number of particles

  • \mathrm{Mole = \frac{W}{M} = \frac{\textrm{(Wt. of substance in gm.) }}{(\textrm{Molar mass of substance (G.m.m)})}}

Here G.m.m. = Gram molecular mass

  • \mathrm{Mole = \frac{(\textrm{Volume of substance in litre}) }{22.4\ litre}}

  • Volume of one mole of any gas is equal to 22.4 litres of dm3 at STP. It is known as molar volume.

  • \mathrm{Mole = \frac{\textrm{Number of identities}}{\textrm{Avogadro's number}}}

  • \mathrm{Mole =\frac{ PV }{RT }}

Here P= Pressure in atmosphere, V = Volume in litre, T = Temperature in Kelvin, R = Universal gas constant



Relationship of Mole:

 A mole of any substance (like O2) stands for: 

  • 6.023 x 1023  molecules of O2 

  • 2 x 6.023 x 1023 atoms of Oxygen 

  • 32 gm of Oxygen

  • 22.4 litre of O2 at STP.


To Find the Total Number of Identities:

  1. Total number of Molecule = mole(n) x NA

  2. Total number of Atoms = mole (n) x NA x No. of atoms present in one molecule 

  3. Total number of Electrons = mole (n) x NA  x No. of electron present in one electron

  4. Total charge on any ion = mole (n) x NA x charge on one ion x 1.6 x 10-19 C


Determination of Molar mass:

  1. Vapour Density method -

  1. Vapour Density of gas is defined as the ratio of the weight of  certain volume of a gas to the weight of the same volume of hydrogen at the same temperature and pressure 

\mathrm{\textrm{Vapour density of a gas} = \frac{\textrm{weight of V litres of the gas}}{\textrm{Weight of V litres of } H_2 \textrm{ at the same temperature and pressure}}}


  1. \mathrm{Vapour\ Density = (Molecular\ mass) / 2 }

  1. Graham Diffusion Method - 

\mathrm{r_1 / r_2 = (M_2/M_1)^{1/2} }

Here r1 ,  r2 are rates of diffusion for two species while M1, M2 are their molecular masses respectively. 

  1. Colligative Properties -

\pi V=\frac{W}{m}RT

Here, \mathrm{\pi = \textrm{Osmotic pressure in atm}}

V = Volume in litre

W = Weight in grams 

R = Universal gas Constant

T = Given Temperature

m = Molar mass

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