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Percent Composition Formula MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

17 Questions around this concept.

Solve by difficulty

MEDIUM

What mass of $95 \%$ pure $\mathrm{CaCO}_{3}$ will be required to neutralise

$50 \mathrm{~mL}$ of $0.5 \mathrm{MHCl}$ solution according to the following reaction?

$\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_\mathrm{{(a q)}} \rightarrow \mathrm{CaCl}_{2(\text { aq })}+\mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}$
[Calculate upto second place of decimal point]

$1.32 \mathrm{~g}$

$3.65 \mathrm{~g}$

$9.50 \mathrm{~g}$

$1.25 \mathrm{~g}$

View Solution

The right option for the mass of $CO_2$ produced by 20 g of 20% pure limestone is (Atomic mass of Ca =40)

$CaCO_3\overset{1200K}{\rightarrow} CaO+CO_2$

1.12g

1.76g

2.64g

1,32g

View Solution

Concepts Covered - 1

Percentage Composition And Equivalent Weight

Percentage Combination:

percentage combination of the compound is the relative mass of the each of the constituent element in 100 parts of it.

$\mathrm{Mass\ \% \ of\ an\ element = \frac{\textrm{Mass of that element in one mole of the compound}}{\textrm{Molar mass of the compound}}\times100}$

Let us take an example of water (H₂O), it contains hydrogen and oxygen, the percentage composition of both these elements can be calculated as follow:

The molar mass of water = 18.02 g

$\mathrm{Mass\ \% \ of\ Hydrogen = \frac{2\times1.008}{18.02}\times100}=11.18 \%$

$\mathrm{Mass\ \% \ of\ Oxygen = \frac{16.00}{18.02}\times100}=88.79 \%$

One can check the purity of a given sample by analyzing percentage composition.

Equivalent Weight:

Equivalent weight is the weight of an element or a compound which combines with or displaces 1.008 part by weight of H₂ or 8 part by weight of O₂, or 35.5 part by weight of Cl₂.
Equivalent weight is a number and when it is
denoted in grams, it is called gram equivalent.
It depends upon the nature of chemical reaction
in which substance takes part

Methods To Find Equivalent Weight:

For Acids:
E = (Molecular weight) / (Protocity or Basicity of Acid)
For Example, for H3PO4, E = M/3
For H₂SO₄ , E =M/2
For Bases:
E = (Molecular weight) / (Acidity or number of OH^- ions)
For Example, for Ca(OH)₂, E = M/2
For Al(OH)₃, E =M/3
For Ions:
E = (Molecular weight) / (Charge on ion)
For Example, for SO₄^2-, E = M/2
For PO₄^3-, E = M/3
For Compounds:
E = (Molecular weight) / (Valency of cation or anion)
For Example, for CaCO₃, E = M/2
For AlCl₃, E =M/3
For Redox Reactions:
E = (Molecular weight) / (Change in oxidation number)
For Example, for KMnO₄
(a) In acidic medium: E = M/5

$\mathrm{\overset{+7}{KMnO_{4}}\, +\, 3H_{2}SO_{4}\, \rightarrow K_{2}SO_{4} + \overset{+2}{2MnSO_{4}} + 3H_{2}O + 5 [O]}$
5 unit change in oxidation number.

(b) In basic medium: E = M/1

$\mathrm{\overset{+7}{2KMnO_{4}}\, +\, 2KOH\, \rightarrow \overset{+6}{2K_{2}MnO_{4}}\, +\, H_2O\, +\,[O]}$
one unit change in oxidation number

(c) In neutral medium: E = M/3

$\mathrm{\overset{+7}{2KMnO_{4}} + H_{2}O\, \rightarrow 2KOH + \overset{+4}{2MnO_{2}} + 3[O]}$
3 unit change in oxidation number
For Acidic Salt:
E = (Molecular weight) / (Number of replaceable H-atoms)
For example, for H₃PO₄