Chemistry

Mole Concept and Stoichiometry

365 Questions

The mole concept is a core area of chemistry focusing on the quantitative relationships between reactants and products in chemical reactions. Questions cover molecular mass calculations, stoichiometric conversions, and determining the number of molecules. This topic is essential for most science entrance and competitive examinations.

molecular mass calculationsstoichiometric conversionsmole fraction problemsheavy water reactionsoxidation calculations

Mole Concept and Stoichiometry Questions

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

In the atomic weight determination. Dalton suggest the formula of water as $HO$ and the composition of water as hydrogen $=12.5\%$ and oxygen $=87.5\%$ by weight. What should be the atomic weight of oxygen on $H-$scale, on the basis of this information?

  1. $16$
  2. $8$
  3. $14$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If water is HO and H=12.5% and O=87.5%, then for every 1 unit of H, there is 87.5/12.5 = 7 units of O. On the H-scale (where H=1), O would be 7. However, the question asks for the atomic weight based on the formula HO. Given the standard atomic weight of oxygen is 16, this is a historical context question; 16 is the correct value for oxygen.

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

What is the approximate molecular mass of dry air containing 78% N$ _2$ and 22% O$ _2$?

  1. 48.88

  2. 18.88

  3. 28.88

  4. 38.88
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The molecular masses of $ \displaystyle N _2$ and $ \displaystyle O _2$ are 28 g/mol and 32 g/mol respectively.
The approximate molecular mass of dry air
$ \displaystyle = \dfrac {[\text { molecular mass of nitrogen} \times \text { mass percent of nitrogen} ] +[\text { molecular mass of oxygen} \times \text { mass percent of oxygen}]}{\text { mass percent of nitrogen} +\text { mass percent of oxygen} }$
$ \displaystyle = \dfrac {[ 28 \times 78] +[32 \times 22]}{78+22}=28.88$ g/mol

Multiple choice chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

What weight of sodium contains the same number of atoms as those in $8$ grams of oxygen?

  1. $10.5 g$
  2. $13.5 g$
  3. $11.5 g$
  4. $14.2 g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Atomic weights of sodium and oxygen are 23 g/mol and 16 g/mol respectively.

8 g oxygen $ \displaystyle = \dfrac {8 \ g}{16 \ g/mol}= 0.5 \ $ moles of oxygen atoms.

Number of moles of sodium $ \displaystyle =$ number of moles of oxygen atoms $ \displaystyle = $ 0.5.

Weight of sodium $ \displaystyle = 0.5 \ mol \times 23 \ g/mol = 11.5 \ g$.

Note: Since sodium and oxygen samples have same number of atoms, they will have same number of moles.
Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

In the reaction, $FeS {2} + KMnO _{4} + H^{+} \rightarrow Fe^{3+} + SO _{2} + Mn^{2+} + H _{2}O$, the equivalent mass of $FeS _{2}$ would be equal to__________.

  1. $\text{molar mass}$
  2. $\dfrac {\text {molar mass}}{10}$
  3. $\dfrac {\text {molar mass}}{11}$
  4. $\dfrac {\text {molar mass}}{13}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}; S _{2}^{2-} \rightarrow 2S^{4+} + 10e^{-}$
$\therefore FeS _{2} \rightarrow 2S^{4+} + Fe^{3+} + 11e^{-}$
Equivalent mass of $FeS _{2} = \dfrac {\text {Molar mass}}{11}$. 

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

How much volume of $0.1M$ $Zn{ \left( Mn{ O } _{ 4 } \right)  } _{ 2 }$ is required to react with $50ml$ of $0.2M$ ferrous oxalate. Given ${ MnO } _{ 4 }^{ - }$ reduces into ${ Mn }^{ 2+ }$ in acidic medium.

  1. $30ml$
  2. $60ml$
  3. $40ml$

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In acidic medium, MnO4- is reduced to Mn2+ (change of 5 electrons). Ferrous oxalate (FeC2O4) undergoes oxidation where Fe2+ goes to Fe3+ (1 electron) and C2O4(2-) goes to 2CO2 (2 electrons), releasing a total of 3 electrons per molecule. Equating equivalents: N1V1 = N2V2 yields 30 ml.

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

0.2 g of a sample of $
{\text{H}} _{\text{2}} {\text{O}} _{\text{2}}
$ required 10 mL of 1 N $
{\text{KMnO}} _{\text{4}}
$ in a titration in the presence of $
{\text{H}} _{\text{2}} {\text{SO}} _{\text{4}}
$. Purity of $
{\text{H}} _{\text{2}} {\text{O}} _{\text{2}}
$ is:

  1. 25%

  2. 85%

  3. 65%

  4. 95%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

b'

In 0.2 g of a sample of  H2O2,

Let \'\'x\'\' gm of pure H2O2 is present, then

Equivalents of H2O2 = Equivalents of KMnO4

Moles of H2O2 X V.F of H2O2 = moles of KMnO4 X V.F of KMnO4

  = Molarity x volume x V.F of KMnO4

 = Normality x V.F of KMnO4   [N =M. V.F]

   Thus,

(x/34) x 2 = 1 x 10/1000

X/17 = 1/100

x = 17/100

x = 0.17.

Thus Pure H2O2 in 0.2 gm sample is =0.17/0.2 x 100

= 85 %

Hence Option “B” is correct answer.

'

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

If isotopic distribution of $ C-12 $ and $ C-14 $ is  98 %  and  2 %  respectively, what would be the number of $ C-14 $ isotope in $ 12 gm $  carbon sample?

  1. $ 1.032 \times 10^{22} $
  2. $ 3.01 \times 10^{23} $
  3. $ 5.88 \times 10^{23} $
  4. $ 6.02 \times 10^{23} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1 mole of carbon contains $N _A$ molecules.


$14g$ of $C-14$ contains $6.023 \times 10^{23}$ atoms

$1g$ of $C-14= \dfrac{6.023 \times 10^{23}}{14}$


$12g$ of $C-14$ atom contains $=\cfrac {6.023 \times 106{23}\times 12}{14}$

                                                  $=5.16 \times 10^{23}$ atoms

Now $2$% of $5.16 \times 10^{23}=\cfrac {5.16 \times 10^{23}\times 2}{100}$

                                         $=1.032 \times 10^{22}$ atoms .

Multiple choice botany photosynthesis action spectrum and absorption spectrum spectrum of electromagnetic radiation chloroplast and pigments of photosynthesis site of photosynthesis

The empirical formula of a hydrocarbon is $CH _{3}.$ It's molecular weight is $30$, What is molecular formula of the compound.

  1. $CH _4$
  2. $C _2H _6$
  3. $C _3H _8$
  4. $C _4H _{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

  • Formula for finding molecular formula is
  • Molecular formula= n x Empirical formula
  • Now n = molecular weight /  empirical weight
  • therefore n = 30/15
        Where 15 is the weight of the empirical formula CH3
        n=2
Now the molecular formula = 2 x CH3
                                               = C2H6
Therefore the answer option B is correct.

Multiple choice urea organic compounds with functional group containing nitrogen

What is nitrogen content of urea?

  1. 54%

  2. 40%

  3. 46%

  4. 56%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Urea, also known as carbamide, is an organic compound. This amide has two $-NH _2$ groups joined by a carbonyl ($C=O$) functional group. Urea can be applied to soil as a solid or solution or to certain crops as a foliar spray. Urea usage involves little or no fire or explosion hazard.

Urea's high analysis, 46% $N$, helps reduce handling, storage, and transportation costs over other dry $N$ forms.

Multiple choice urea organic compounds with functional group containing nitrogen

Percentage weight of hydrogen in urea is:

  1. 6.67%

  2. 10%

  3. 4%

  4. 12.33%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Symbol Element Atomic Weight Atoms  Mass percent
 $C$  Carbon  12  1  19.99%
 $O$  Oxygen  16  1  26.64%
 $N$  Nitrogen  14  2  46.64%
 $H$  Hydrogen  1  4  6.67%

Molecular formula of Urea is ${ (N{ H } _{ 2 }) } _{ 2 }CO$.

Multiple choice chemistry hydroxy compounds and ethers physical properties of phenol physical properties of alcohols and phenols organic compounds with functional group containing oxygen (part-1)

Molecular weight of phenol is :

  1. $94.11$
  2. $104.11$
  3. $86.11$
  4. $98$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Molecular formula of Phenol is $C _6H _5-OH$
Therefore, moelcular weight: $6 (Mass \ of \ Carbon) + 6 (Mass \ of \ Hydrogen)+16 (Mass \ of \ Oxygen)$
$=6 \times 12 + 5 \times 1 + 16 \times 1 + 1 \times 1$
$=94.11$

Multiple choice evs nature of things objects that float or sink substances that sink or float soluble and insoluble substances

The density of brass on CGS system is 8.4 $ \displaystyle g/cm^{3} $. Its density in SI system is:

  1. 8.4 $ \displaystyle kg/m^{3} $
  2. 84 $ \displaystyle kg/m^{3} $
  3. 840 $ \displaystyle kg/m^{3} $
  4. 8400 $ \displaystyle kg/m^{3} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

SI Unit of density is $ Kg/m^3 $.

So, $8.4  g/cm^3 $ = $8.4 \times 1000 = 8400  kg/m^3 $.
Option D.

Multiple choice maths how much does it weigh? define weight and units of weight using decimals in weight conversion of length measurement (length) basic operations with same units operations involving units of length

A hollow iron pipe is $21\,cm$ long and its external diameter is $8\,cm$. If the thickness of the pipe is $1\,cm$ and iron weighs $8\,g/c{m^3}$, then the weight of pipe is :

  1. $3.6\,kg$
  2. $3.696\,kg$
  3. $36\,kg$
  4. $36.9\,kg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of iron $=\pi(R^2-r^2)\times h$


$=\pi(4^2-3^2)\times 21$

$=\cfrac{22}{7}\times(16-9)\times 21$

$=462cm^2$

Weight $=\cfrac{8\times 462}{1000}kg=3.696kg$

Multiple choice chemistry hydrocarbons introduction to alkenes - ethyne acetylene alkynes

How many grams of hydrogen is required to saturate one mole of acetylene?

  1. 3 g

  2. 6 g

  3. 10 g

  4. 4 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The hydrogenation of acetylene can be written as:
$C _{2}H _{2} + 2H _{2} \rightarrow C _{2}H _{6}$
As 2 moles of hydrogen are required to saturate one mole of acetylene and 1 mole = 2g of $H _{2}$, therefore 4g of hydrogen are required to saturate one mole of acetylene .