28 Year NEET Questions: Some Basic Concepts Of Chemistry - 1

18 Questions MCQ Test Chemistry 31 Years NEET Chapterwise Solved Papers | 28 Year NEET Questions: Some Basic Concepts Of Chemistry - 1

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In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end? [2003]


It is given that only 50% of the expected product is formed.
Hence, only 10 litre of NH3 is formed i.e. N2 used = 5 litres ⇒ N2 left = 30 – 5 = 25 litres
H2 used = 15 litres ⇒ H2 left = 30 – 15 = 15 litres


The maximum number of molecules is present in:


No. of molecules in different cases:

(a) 22.4 litre at STP contains = 6.023×1023 molecule of H2
∴ 15 litre at STP contains molecules

(b) 22.4 litre at STP contains = 6.023×1023 molecule of N2
∴ 5 litre at STP contains molecules

(c) 2 gm of  H= 6.023×1023 molecules of  H2
∴ 0.5 gm of H2 contains molecules

(d) Similarly, 10 g of O2 gasmolecules

Thus (a) will have maximum number of molecules


The mass of carbon anode consumed (givin g only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is  (Atomic mass: Al = 27) [2005]


2Al2O3 + 3C → Al3CO2

Gram equivalent  of Al2O3 = Gram equivalent of C

Equivalent weight of Al 

Equivalent weight of C

No. of gram equivalent of Al  = = 30 × 103
Hence, No. of gram equivalent of C = 30 × 103

Again, No. of gram equivalent of C
⇒  mass = 90 × 10g = 90 kg


The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is:   [2005]



Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL– 1. Volume of acid required to make one litre of 0.1MH2SO4 solution is [2007]


Molarity of H2SO4 solution

Suppose V ml of this H2SO4 is used to prepare 1 lit. of 0.1M H2SO4
∴ V x 18.02 = 1000 x 0.1


An element , X has the following isotopic composition :
200X:90 %
199X : 8.0%
202X ; 2.0 %
The weighted average atomic mass of the naturally occuring element X is closest to


Average isotopic mass of,199.96 amu


The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is [2007]


The balanced chemical equation is:

From the equation, it is clear that 2 moles of MnO4  are required to oxidize 5 moles of SO3–2  
2/5 mole of MnO4 is required to oxidize 1 mole of SO3–2 


Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]


Density of water = 1 gram cm–3 
∴ Volume occupied by 1 gram water = 1 cm3 
i.e. Volume occupied by  molecules of water = 1 cm3 [∵ 1g water = 1/18moles of water]

Thus volume occupied by 1 molecule of water 


Number of moles of MnO4- required to oxidize one mole of ferrous oxalate completely in acidic medium will be : [2008]



What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1L of propane gas (C3H8) measured under the same conditions? [2008]


Writing the equation of  combustion of propane (C3H8), we get

From the above equation we find that we need 5 L of oxygen at NTP to completely burn 1 L of propane at N.T.P.
If we change the conditions for both the gases from N.T.P. to same conditions of temperature and pressure. The same results are obtained. i.e. 5 L is the correct answer.


An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be: [2008]


Thus empirical formula is CH3O.


How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl ? [2008]


Writing the equation for the reaction, we get:

From this equation, we find 223 g of PbO reacts with 73 g of HCl to form 278 g of PbCl2.
If we carry out the reaction between 3.2 g HCl and 6.5 g PbO.
Amount of PbO that reacts with 3.2 g HCl =

Since the amount of PbO present is only 6.5 g so PbO is the limiting reagent.

Amount of PbCl2 formed by 6.5 g of PbO

Number of moles of PbCl2 formed


10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be: 


In this reaction, oxygen is the limiting agent.
Hence the amount of H2O produced depends on the amount of O2 taken
0.5 mole of O2 gives H2O = 1 mol
∴ 2 mole of O2 gives H2O = 4 mol


What is the [OH] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2? [2009]


No. of milli equivalent of HCl = 20 × 0.05 = 1.0
No. of milli equivalent of Br (OH)2   = 30 × 0.1 × 2 = 6.0
After neutralization, no. of milli equivalents in 50 ml. of solution = (6 – 1) = 5
Total volume of the solution = 20 + 30 = 50 ml
∴ No. of milli equivalent of OHis 5 in 50 ml, 


The number of atoms in 0.1 mole of a triatomic gas is: [2010] (NA = 6.02 ×1023 mol–1)


The number of atoms in 0.1 mole of a triatomic gas = 0.1 × 3 × 6.023 × 1023.= 1.806 × 1023


Which has the maximum number of molecules among the following ? [2011 M]


Hence, option C is correct.


6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is: [NEET 2013]



In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO3, which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine): [NEET Kar. 2013]


Millimoles of solution of chloride = 0.05 × 10 = 0.5
Millimoles of AgNO3 solution = 10 × 0.1 = 1
So, the millimoles of AgNO3 are double than the chloride solution

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