RBSE Class 11 Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

These comprehensive RBSE Class 11 Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry will give a brief overview of all the concepts.

RBSE Class 11 Chemistry Chapter 1 Notes Some Basic Concepts of Chemistry

→ Science: Science is the specific knowledge which deals with the systematic and organized study of any subject.

→ Matter: Anything which has mass and occupies space is called matter.

→ Homogeneous Mixture: In this type of mixture, components of mixture are completely mixed with each other and its composition is uniform throughout.

→ Heterogeneous Mixture: In this mixture, the composition of components of mixture is not uniform throughout. In this, the components are separately seen.

→ Accuracy: It is the difference between the experimental value or the mean value of a set of measurement and true value.

→ Precision: It is expressed as difference between a measured value and arithmetic mean value from a series of measurements.

→ Significant Figure: In the measured value of physical quantity, the digits about the correctness of which we are surplus the last digit which is doubtful, are called significant figures.

→ Law of Conservation of Mass: According to this law, “matter can neither be created nor destroyed”.

→ Law of Constant Composition: According to this law, “the ratio of mass of elements in any compound remains same, whatever method is used to prepare the compound.”

 

→ Law of Multiple Proportions: This law states that “when two elements combine to form more than one compound, then the mass of one of the elements which combine with fixed mass of the other element are in a simple whole number ratio.”

→ Law of Reciprocal Proportions: According to this law, “when two elements combine separately with a fixed mass of a third element, then the ratio of their masses in which they do so in either the same or same whole number multiple of the ratio in which they combine to each other.

→ Gay-Lussac’s Law of Combining Volumes: When gases react with each other, they do so in volumes which bear a simple whole number ratio to one another and to the volume of gaseous products

→ Avogadro Law: At normal temperature and pressure, equal volumes of all gases contain equal number of molecules.

→ Atomic Mass: Atomic mass is defined as mass exactly equal to the ~th parts of mass of isotope of carbon-12 atom.

→ Equivalent Weight: The number of parts by mass of a substance that combines with or displaces 1.008 part by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine is called equivalent weight.

→ All fion zero digits are significant numbers.

→ If a number less than one, all zero to the right of decimal point are significant where as left side number of decimal are not significant.

→ There are seven base physical quantities in S.I. system.

→ Atomic mass unit or a.m.u. or u = Mass of \(\frac{1}{12}\)th part of C-12 isotope.

→ 1 mole = 6.022 × 10²³ molecules

→ Gram atomic mass = 6.022 × 10²³ atom’s mass (g)

→ Gram molar mass =6.022 × 10²³ molecules

→ Molecular Mass =2 × Vapour density

→ M = \(\frac{W_B \times 1000}{M_B \times V_{(\mathrm{mL})}}\)

→ Molecular formula = n × Empirical formula

→ m = \(\frac{W_B \times 1000}{M_B \times W_A}\)

→ Molarity = \(\frac{M_1 V_1}{n_1}=\frac{M_2 V_2}{n_2}\)

→ Normality N₁V₁ = N₂V₂

→ Approximate Atomic Mass = \(\frac{6.4}{\text { Specific Heat }}\)

→ Valency =\(\frac{\text { Approx. Atomic Mass }}{\text { Equivalent Mass }}\)

→ Equivalent Mass =\(\frac{\text { Atomic mass }}{\text { Valency }}\)

→ Equivalent Weight of Acid =\(\frac{\text { Molecular Mass }}{\text { Basicity }}\)

→ Equivalent Weight of Base = \(\frac{\text { Molecular Mass }}{\text { Acidity }}\)

→ Equivalent Weight of = \(\frac{\text { Molecular Mass }}{\text { Total Positive Charge }}\)

→ x_A = \(\frac{n_A}{n_A+n_B}\), x_B = \(\frac{n_B}{n_A+n_B}\)

→ x_A + x_B = 1

→ Molarity(M) = \(\frac{M_1 V_1+M_2 V_2}{V_1+V_2}\)

→ m = \(\frac{1000 M}{1000 \mathrm{~d}-M \cdot M_B}\)

→ M = \(\frac{p \times d \times 10}{\text { Molecular weight of solute }}\)

→ N = \(\frac{p \times d \times 10}{\text { Equivalent weight of solute }}\)

→ % mole =\(\frac{\text { Moles of component }}{\text { Total moles of mixture }}\) × 100

→ X_B =\(\frac{m M_B}{100-m M_B}\)

→ Atomic weight = Equivalent weight × Valency

→ Molecular weight = 2 × Vapour density

→ Dulong and Patit’s Law ⇒ Atomic weight × Specifi heat = 6.4 (Approx)

→ X_B =\(\frac{M}{M+\frac{1000 \mathrm{~d}-M M_B}{M_A}}\)

→ N =\(\frac{W_B+1000}{(\mathrm{Eq} \cdot \mathrm{Mass})_B \times V_{(\mathrm{mL})}}\)

→ No. of atoms in W g of element = \(\frac{W \times 6.022 \times 10^{23}}{\text { Gram atomic weight }}\)

→ No. of moles in x molecule =\(\frac{x}{6.022 \times 10^{23}}\)

→ No. of mole in n molecule = n × 6.023 × 10²³

→ No. of moles in w(g) = \(\frac{w}{\text { Gram molecular weight }}\)

→ Volume of n mole of gas at N.T.P. = n × 22.4 L