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An organic compound contains 80% (by wt.) carbon and the remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12, H is 1] (2021)
Hence empirical formula =CH3
Additional Information: An empirical formula represents the simplest whole-number ratio of various atoms present in a compound, whereas, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.
Which one of the followings has maximum number of atoms? (2020)
We know that,
Number of atoms =
(a) no. of atoms
(b) no. of atoms
(c) no. of atoms
(d) no. of atoms
∴ 1 gm of Li is having maximum no. of atoms.
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
6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is: [NEET 2013]
Given, the number of molecules of urea =6.02 x 1020
therefore, the number of moles =
Therefore, the correct answer is 0.01 M
Which has the maximum number of molecules among the following ? [2011 M]
Hence, option C is correct.
The number of atoms in 0.1 mole of a triatomic gas is:  (NA = 6.02 ×1023 mol–1)
1 mol of triatomic gas has 3 ×6.02×1023 atoms
Therefore, no. of atoms in 0.1 mol =0.1× 3 ×6.02×1023 = 1.806×1023
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 = 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 OH– is 5 in 50 ml,
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 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
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: 
The empirical formula is to be calculated as:
Thus empirical formula is CH3O.
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.
Number of moles of MnO4- required to oxidize one mole of ferrous oxalate completely in acidic medium will be : 
No. of moles of MnO4− required to oxidize one mole of FeC2O4 completely in an acidic medium.
3MnO4− + 24H+ + 5FeC2O4⟶ 3Mn+2 + 12H2O + 10CO2 + 5Fe+3
5 moles of FeC2O4 reacts with 3 moles of MnO4−.
5 moles of FeC2O4⟶3 moles of MnO4−
1 mole of FeC2O4⟶ 53 moles of MnO4−
∴ The number of moles of MnO4− required to oxidize one mole of ferrous oxalate completely in the acidic medium will be 0.6 moles.
Volume occupied by one molecule of water (density = 1 g cm–3) is : 
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
The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is 
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
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
The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is: 
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 = 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 × 103 g = 90 kg
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 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:
As we know, 1 mole of gas has NA molecules. So, the higher the number of moles of a gas, it will have higher the number of molecules.
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 H2 = 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
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,
= 199.96 amu ≈200 amu
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:
2H2 + O2 → 2H2O
10 g of hydrogen (molar mass 2 g/mol) = 2 g/mol 10 g = 5 mol
64 of oxygen (molar mass 32 g/mol) = 32 g/mol 64 g = 2 mol
As per the balanced chemical equation, 1 mole of oxygen will react with 2 moles of hydrogen to form 2 moles of water.
Here, oxygen is the limiting reagent and hydrogen is the excess reagent.
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