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 acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

A 26 ml of $ N-Na _{2}CO _{3} $ solution is neutralized by the solutions of acids A and B in different experiments. The volumes of the acids A and B required were $10 ml$ and $40 ml$, respectively. How many volumes of A and B are to be mixed in order to prepare 1 litre of normal acid solution? 

  1. $179.4, 820.6$
  2. $820.6, 179.4$
  3. $500, 500$
  4. $474.3, 525.7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2} (A)$
$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2}(B)$
$ N _{1} = 1 , V _{1} = 26ml, V _{2} = 10ml, V _{3} = 40ml$
Normality of A
$ N _{2} = \dfrac{N _{1}V _{1}}{V _{2}} = \dfrac{1 \times 26}{10} = 2.6 N $
Normality of B
$ N _{3}= \dfrac{N _{1}V _{1}}{V _{3}} = \dfrac{1 \times 26}{40} = 0.65 N $
if we mix A & B then
$ N _{1}V _{1} + N _{2} V _{2} = N _{3} (V _{1}+V _{2})$
$ 2.6 V _{1}+0.65V _{2} = 1 \times 1000 $
$ 2.6V _{1}+ V _{2} + 0.65 V _{2} = 1000 ...(1)$
$ V _{1}+V _{2} =1000 ...(2)$
multiply $eq^{n}$ (2) by 0.65 
$ 0.65 V _{1}+0.65V _{2} = 650 ...(3)$
$ V _{2} = 820.6 ml$
$ V _{1} = 1000 - 820.6 = 179.4 ml $
option "a" correct.

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

What volume of $ 0.18 N - KMnO _{4} $ solution would be needed for complete reaction with 25 ml of $ 0.21 N - KNO _{2} $ in acidic medium ?

  1. $57.29 ml$
  2. $11.67 ml$
  3. $29.17 ml$
  4. $22.92 ml$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ KMnO _{4}+KNO _{2}$
$ 0.18 N $ 25 ml
V = ? 0.21 N
we know $ N _{1}V _{1}= N _{2}V _{2}$
$ 0.18 \times V _{1} = 0.21 \times 25 $
$ V _{1} = \dfrac{0.21 \times 25}{0.18 }$
$ \boxed{V _{1} = 29.17 ml}$
Answer option C

Multiple choice chemistry structure of the atom neutrons discovery of neutrons structure of atoms

Number of neutrons present in $1.7$ g of ammonia is:

  1. $0.8N _A$
  2. $0.7N _A$
  3. $0.652N _A$
  4. $0.52N _A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let us consider the problem.
For, moles of $N{H _3} = \frac{{1.7}}{{17}} = 0.1$
Let the Neutrons in 1 molecules of $N{H _3} = 7 + 3 \times 0 = 7$
Hence,
Number of Neurons $= 0.1 \times {N _A} \times 7$
The correct answer is $0.7{N _A}$
Multiple choice chemistry structure of the atom neutrons discovery of neutrons structure of atoms

Calculate number of neutrons present in $12 \times 10^{25}$ atoms of oxygen $( _8O^{17})$:
(Given : N$ _A = 6 \times 10^{23}$)

  1. $1800$
  2. $1600$
  3. $1800\ N _A$
  4. $3200\ N _A$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Atomic number =$ P =8$ 


Atomic Mass =$ N+P=17$


                         $N=9$
 Total no. of Neutrons = $9\times 12\times 10^{25}$

                                     =$\dfrac{9\times 12\times 10^{25}\times N _{A}}{6\times 10^{23}}$ 

                                 $=1800N _{A}$

Hence, the correct option is C.

Multiple choice chemistry medicinal chemistry antacids and antihistamines therapeutic action of different classes of drugs ph regulation of the stomach

An antacid tablet weighing 1 g containing aluminium hydroxide as the only basic substance and the rest of its components being neutral, was dissolved in 200 mL of 0.1 M HCI. The excess HCI was back titrated and required  90 mL of 0.1 N base for exact neutralization. Mill equivalents of aluminum hydroxide in the sample of antacid tablet is 

  1. 9

  2. 11

  3. 12

  4. 20
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total milliequivalents of HCl = 200 mL * 0.1 M = 20 meq. The excess HCl neutralized by 90 mL of 0.1 N base = 9 meq. Therefore, the HCl consumed by the antacid = 20 - 9 = 11 meq. Since aluminum hydroxide is the only base, it must have neutralized 11 meq of HCl.

Multiple choice chemistry introduction to analytical chemistry different types of solutions various mixtures introduction to solutions

What volume would you dilute 0.2 L of a 15 M solution to obtain a 3 M solution?

  1. 1L

  2. 225L

  3. 10L

  4. 0.4L

  5. 0.1L
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that :

$\displaystyle M _1V _1 = M _2V _2 $ 

Given :

$\displaystyle  M _1 = 15 $  M
$\displaystyle V _1 = 0.2 $ L
$\displaystyle  M _2 =3$  M
$\displaystyle  V _2 =$  ????

$\displaystyle  15 \times 0.2 = 3 V _2$
Thus, $\displaystyle  V _2 = 1$  L

Hence, the correct option is A.
Multiple choice chemistry separation of substances classification of mixtures mixtures: examples and properties types of solutions

$3$ litre of mixture of propane $(C {3}H _{8})$ $ butane $(C{4}H_{10})$ on complete combustion gives $10$ litre $CO_{2}$. Find the composition of mixture.

  1. $C _{3}H _{8}2L$ and $C _{4}H _{10}\ 1L$
  2. $C _{3}H _{8}3L$ and $C _{4}H _{10}\ 0L$
  3. $C _{3}H _{8}\ 1.5L$ and $C _{4}H _{10}\ 1.5L$
  4. $C _{3}H _{8}\ 0L$ and $C _{4}H _{10}\ 3L$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x be the volume of propane and y be the volume of butane. We have x + y = 3. The combustion reactions are C3H8 + 5O2 -> 3CO2 + 4H2O and C4H10 + 6.5O2 -> 4CO2 + 5H2O. Thus, 3x + 4y = 10. Solving these equations gives x = 2 and y = 1.

Multiple choice chemistry endothermic and exothermic reactions endothermic reactions what are enthalpy changes energy change in chemical reactions
For per gram reactant, the maximum quantity of ${ N } _{ 2 }$ gas is produced in which of the following thermal decomposition reaction? (Given: Atomic wt. Cr=52 u, Ba=137 u)
  1. $(NH _4) _2Cr _2O _7(s) \rightarrow N _2(g) + 4H _2O(g)+Cr _2O _3(s)$
  2. $2NH _4NO _3(s) \rightarrow 2N _2(g) + 4H _2O(g)+O _2(g)$
  3. $Ba(N _3) _2(s) \rightarrow Ba(s) + 3N _2(g)$
  4. $2NH _3(g) \rightarrow N _2(g) + 3H _2(g)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The % composition of four hydrocarbons are as follows:


(i) (ii) (iii) (iv)
%C 75 80 85.7 91.3
%H 25 20 14.3 8.7


The data illustrates the law of:

  1. constant proportion

  2. conservation of mass

  3. multiple proportion

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

The law of multiple proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. This table shows different hydrocarbons, illustrating this law.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The oxides of nitrogen contain 63.65%, 46.69% and 30.46% of nitrogen by weight, respectively. This data illustrates the law of:

  1. constant proportions

  2. multiple proportions

  3. reciprocal proportions

  4. conservation of mass
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The law of multiple proportions say that If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.
Here, the ratios of oxygen which react with a constant mass of N are $2:3 , 2:1$ and $4:3$ respectively.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

0.75 mole of solid ‘$A _4$’ and 2 moles of gaseous $O _2$ are heated in a sealed vessel, completely using up the reactants and producing only one compound. It is found that when the temperature is reduced to the initial temperature, the contents of the vessel exhibit a pressure equal to half the original pressure. What conclusions can be drawn from these data the product of the reaction?

  1. $A _2O _5$
  2. $A _3O _4$
  3. $A _2O _2$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
give reaction
$A _4(s) + O _2(g) → A _xO _y$

Applying POAC  law for A atoms,
4 × moles of $A _4  =  x \times moles of A _xO _y$
=> $4 \times 0.75 = x \times  1$
=> x = 3

Applying POAC for O atoms,
$2\times  moles of O _2 = y \times moles of A _xO _y$
=> $2 \times 2 = y \times 1$
=> y = 4

Thus, the formula of the product is $A _3O _4$.
Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Two samples of lead oxide were separately reduced to metallic lead by heating in hydrogen. The percentage weight of lead from one oxide was half the percentage weight of lead obtained from the other oxide. The data illustrates the law of:

  1. reciprocal proportions

  2. constant proportions

  3. multiple proportions

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

The law of multiple proportions say that If two elements form more than one compounds between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.