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? 
It is given that only 50% of the expected product is formed hence only 10 litre of NH3 is formed N2 used = 5 litres, left = 30 – 5 = 25 litres H2 used = 15 litres, 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 = 6.023×1023molecule of H2
∴ 15 litre at STP contains
(b) 22.4 litre at STP contains
= 6.023×1023 molecule of N2
5 litre at STP contians
(c) 2 gm of H2= 6.023×1023 molecules of H2 0.5 gm of H2
(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) 
2Al2O3 + 3C → Al3CO2
Gram equivalent of Al2O3 gm equivalent of C
Now equivalent weight of Al
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 × 103g = 90 kg
The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is: 
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 
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 :
199X : 8.0%
202X ; 2.0 %
The weighted average atomic mass of the naturally occuring element X is closest to
Average isotopic mass of
The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is 
The balance chemical equation is :
From the equation it is clear that Moles of MnO4– require to oxidise 5 moles of SO3– – are 2
Moles of MnO4– require to oxidise 1 mole of SO3– – are 2/5.
Volume occupied by one molecule of water (density = 1 g cm–3) is : 
1 gram cm–3
∴ Volume occupied by 1 gram water = 1 cm3 or Volume occupied by
molecules of water = 1 cm3
[∴ 1g water = moles of water]
Thus volume occupied by 1 molecules of water
i.e. the correct answer is option (c).
Number of moles of MnO4- required to oxidize one mole of ferrous oxalate completely in acidic medium will be : 
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? 
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: 
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 ? 
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 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 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? 
No. of milli equivalent of HCl = 20 × 0.05
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 OH– is 5 in 50 ml
The number of atoms in 0.1 mol of a triatomic gas is :(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]
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