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Some Basic Concepts of Chemistry

MOLE

1 mole of any substance contains 6.023 $\times$ 10²³ particles and this number is known as Avogadro Number.

Avogadro Number(N_A) = 6.023 $\times$ 10²³

$∴$ 1 mole = 6.023 $\times$ 10²³ particles

Note : This particles can be electron, proton, newtron, atom, ion, molecule etc.

Atomic Mass and Gram Atomic Mass

Mass of one atom of any element is called Atomic Mass, It's unit is a.m.u (atomic mass unit).

1 a.m.u = $\frac{1}{N _{A}} g m = 1.67 \times 1 0^{- 27} k g or 1.67 \times 1 0^{- 24} g m$

When we express mass of one mole of atoms in grams then it is called gram atomic mass (GAM) or gram atom or mole atom.

Average Atomic Mass(AAM) :

Definition : Weighted average of atomic mass of isotopes of naturally occuring elements known as Average Atomic Mass(AAM).

AAM = $\frac{\sum ( % o f I so t o p es \times A t o mi c M a ss )}{\sum ( % o f I so t o p es )}$

Number of Gram Molecules

Number of Gram Molecules means Number of Moles of a substance.

Number of Gram Molecules or Number of Moles = $${{Given\,Mass} \over {GMM}}$$

Avogadro's Hypothesis for Gases

Statement : Equal volumes of all gases contains equal number of moles under similar condition of pressure and temperature.

V $\propto$ n at constant P and T

where V = Volume, n = Number of moles

% by Volume

% by volume : - (% v / v)
Volume of Solute (in ml) present in per hundred ml of solution is called % by volume (% v/v).

% $v / v = \frac{v _{so l u t e}}{v _{so l u t i o n}} \times 100$

v_solution = v_solvent + v_solute

% wt by volume :- (% w/v)

% wt by volume :- (% w/v) Weight of solute present per hundred ml of solution is known as % wt by volume.

% $w / v = \frac{w _{so l u t e}}{v _{so l u t i o n}} \times 100$

V_solution = V_solute + V_solvent

Molarity

Molarity : - (M)

* It is denoted by capital M.

* Molarity of solution is defined as number of moles of solute present per litre of solution.

$M = \frac{n o . o f m o l es ( so l u t e )}{v ( in l i t re o f so l u t i o n )}$

$∴$ n = M $\times$ v(in litre) (Applicable for aqueous solution)

where n = Number of moles of solute

$∙$ W gm of a solute having molar mass m₁ is dissolved in sufficient water to form v ml of solution. Determine derived formula of molarity.

$\Rightarrow$ $n_{so l u t e} = \frac{w}{m _{1}}$

$M = \frac{\frac{w}{m _{1}}}{\frac{v}{1000}}$

$= \frac{w}{m _{1}} \times \frac{1000}{v}$

$= \frac{1000 w}{m _{1} v}$

** By default volume of solution is nothing but volume of solvent [V_solution $≅$ V_solvent]

** There is no change in volume of solution on dissolution of solute.

% by weight :- (% w/w)

% by weight :- (% w/w) Weight of solute present in 100 gm of solution is defined by as % by weight.

% $w / w = \frac{w _{so l u t e}}{w _{so l u t i o n}} \times 100$

V_solution = V_solute + V_solvent

Mole Fraction

It is denoted by x and defined as ratio of number of moles of substance to the total number of moles.

Mole fraction (x) =

\frac{n _{s u b s t an ce}}{n _{t o t a l}}

Let a solution contains Solvent(A) and Solute(B) then

Mole fraction of solvent A = x_A =

\frac{n _{A}}{n _{A} + n _{B}}

and Mole fraction of solute B = x_B =

\frac{n _{B}}{n _{A} + n _{B}}

x_A + x_B = 1

Molality

Molality : - (m)

* It is denoted by small m.

* Molality of solution is defined as number of moles of solute present per kg of solvent.

$m = \frac{n o . o f m o l es ( so l u t e )}{w e i g h t ( in k g ) o f so l v e n t}$

Parts per Million (ppm)

Parts per $(10^{6}) mi ll i o n$ (ppm) :

It is a special concentration term which is used when solute present in the solution is negligible then we use ppm.

It is defined as mass of solute present per million that is 10⁶ gm of solution.

$∴$ $pp m = \frac{w _{so l u t e}}{w _{so l u t i o n}} \times 1 0^{6}$

$pp b = \frac{w _{so l u t e}}{w _{so l u t i o n}} \times 1 0^{9}$

ppb = parts per billion

It is used for hardness of water.

Dilution Formula :

$∙$ Process of addition of water in any solution is known as dilution.

n_solute = Constant

V_solution = Increase

M_solution = decrease

Let for a solution initial volume is V₁ and initial molarity is M₁, after adding water of volume V_w in the solution volume becomes V₂ (V₂ = V₁ + V_w) and molarity becomes M₂.

As volume solute stays constant so moles of solute also stay constant.

$∴$ Moles of solute before dilution = Moles of solute after dilution

(n_solute)_BD = (n_solute)_AD [As n = M $\times$ V]

M₁V₁ = M₂V₂

Molarity of Mixing

$$\bullet$$ It is applicable for more than one solution.

$$\left. {\matrix{ {(i)\,Same\,substan ce\,(HCl - HCl)} \cr {(ii)Same\,ion\,(NaCl - NaN{O_3})} \cr } } \right]$$ No reaction

$$\bullet$$ It is not applicable for substance having different nature where they can react.

Eq. $$\matrix{ {\mathop {HCl}\limits_A } & {\mathop {NaOH}\limits_B } \cr } $$ react with each other.

Let a M₁ molar solution of volume V₁ liter is mixed with a M₂ molar solution of volume V₂ liter. Then molarity of the mixture is

$${M_R} = {{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$$

$$\therefore$$ $${M_R} = {{\sum {{M_i}{V_i}} } \over {\sum {{V_i}} }}$$

Special Formulas :

(1) $M = \frac{10 \times % w / v}{M _{so l u t e}}$ (Relation between M and % w/v).

(2) $M = \frac{10 \times % w / w \times d}{M _{so l u t e}}$ (Relation between M and % w/w).

(3) % w/v = % w/w $\times$ d (Relation between % w/v and % w/w).

(4) $m = \frac{x _{so l u t e}}{x _{so l v e n t}} \times \frac{1000}{M _{so l v e n t}}$ (Relation between m and x)

(5) $m = \frac{1000 \times M}{1000 \times d - M \times M _{so l u t e}}$ (Relation between M and m).