NEET Chemistry Questions: Some Basic Concepts in Chemistry

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Assuming fully decomposed, the volume of CO2 released at STP on heating 9.85g of BaCO3 [at. mass of Ba = 137] will be





No explanation available.

The total number of valence electrons in 4.2g of N3- ion is (NA is the Avogadro's number)





(c) Moles of N3- ion = 4.2/42=0.1

Each nitrogen atom has 5 valence electrons. Therefore, total number of electrons of N3- ion=16

Total number of electrons in 0.1 mole or 

4.2g of N3- ion= 0.1x16xNA 

=1.6NA 

In Haber process 30L of  dihydrogen and 30L of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gasesous mixture under the a fore said condition in the end?





As only 50% of the expected product is formed, hence only 10L of NH3 is formed.

Thus, for the production of 10L of NH3 , 5L of  N2 and 15L of H2 are used and composition of gaseous mixture under the a fore said condition in the end is 

H2 = 30-15=15L

N2 = 30-5=25L

NH3 =10L

 

The number of atoms in 4.25 g of NH3 is approximately





(D) Weight of NH3 = 4.25g

Number of moles of NH3 = WeightMolecular weight=4.2517=0.25 mol

Number of molecules in 0.25 mole of NH3 = 0.25 x 6.023 x 1023

So, number of atoms = 4 x 0.25 x 6.023 x 1023

=6.0 x 1023

In the reaction, 4NH3(g)+ 5O2(g) 4NO(g) +6H2O(l)

When 1 mole of ammonia and 1 mole of O2 are made to react to completion, then





(c) 4NH3(g)+ 5O2(g) 4NO(g) +6H2O(l)

      4mol         5 mol       4mol        6mol

According to the equation,

5 moles of O2 required = 4 moles of NH3

1 mole of O2 required =4/5= 0.8 moles of NH3

While 1mole of NH3 requires = 5/4=1.25 moles of O2 

As there is 1 mole of NH3 and 1 mole of O2 , so all the oxygen will be consumed.

An element, X has the following isotopic composition:

200X : 90%, 199X : 8.0%, 202X : 2.0%

The weighted average atomic mass of the naturally occurring element X is closest to 





(d) Weight of 200X = 0.90 x 200 = 180.00 u

Weight of 199X =0.08 x 199=15.92 u

Weight of 202X = 0.02 x 202=4.04 u 

Total weight = 199.96 = 200u

The hydrated salt Na2SO4.nH2O, undergoes 55% loss in  mass on heating and becomes anhydrous. The value of n will be:






Na2SO4.10H2O

12 g of Mg (atomic mass=24) will react completely with an acid to give:





(b) Mg + 2HCl MgCl2 + H2

Therefore, 

12g Mg gives 1/2 mole of H2

The molality of 15% (w/vol.) solution of H2SO4 of density 1.1g/cm3 is approximately:





(d) m = 1598×(100 ×1.1-15)1000=1.6

H3PO4 is a tribasic acid and one of its salts is NaH2PO4. What volume of 1 M NaOH should be added to 12g NaH2PO4 (Molar mass 120) to exactly convert it into Na3PO4?





(c) Meq. of NaH2PO4 =Meq. of NaOH;

Thus, 12/120/2*1000 = 1 x V

V = 200 mL

Which mode of expressing concentration is independent of temperature?





(d) The terms which involves only masses in their formula 

[e.g., molality = mass of solute x 1000 / molar mass of solute x mass of solvent]

are independent of temperature. On the other hand, since, volume changes with temperature, the terms having volume in their formula [e.g., molarity = mass of solute x 1000 / molar mass of solute x volume of solvent],NORMALITY  are dependent on temperatue. 

The product of atomic mass and specific heat of a metal is approximately 6.4. This was given by:





(d) Dulong Petit's law: atomic mass x sp. heat =6.4

0.5 g of fuming H2SO4 (oleum) is diluted  with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is:





(c) Meq. of H2SO4 + Meq.Of SO3 = Meq. of NaOH

(0.5-a)/49*1000 +a/40*1000 = 26.7 x 0.4

a = 0.103

% of SO3 = 0.103/0.5*100 = 20.6%

A metal M forms a compound M2HPO4. The formula of the metal suphate is:





(a)  M2HPO4 means valence of metal is one and thus, sulphate of metal is M2SO4.

A partially dried clay mineral contains 8% water. The original sample contained 12% water and 45% silica. The% of silica  in the partially dried sample is nearly:





(d) Silica    Water    Clay        Mineral

      45         12         43         Initial %

      a            8        (92-a)      % after heating             

 

( in dry state ,8% loss of water is there ,so remaining weight is 100 - 8 = 92)

The % ratio of silica and clay remains constant on heating 

i.e., 4543=a92-a

a = 47%