Relative atomic mass (R.A.M)

$$R.A.M =\frac{\text { Mass of one atom of an element }}{\frac{1}{12} \times \text { mass of one carbon atom }}$$

$$ = \text {Total Number of nucleons}$$

Density:

$$ \text { Specific gravity }=\frac{\text { density of the substance }}{\text { density of water at } 4^{\circ} \mathrm{C}} $$

For gases :

$$ \text {Absolute density (mass/volume)} =\frac{\text { Molar mass of the gas }}{\text { Molar volume of the gas }}$$

$$\Rightarrow \rho=\frac{\mathrm{PM}}{\mathrm{RT}}$$

$$\text {Vapour density V.D.} =\frac{d_{\text {gas }}}{d_{H_{2}}}=\frac{P M_{\text {gas } / R T}}{P_{H_{2} / R T}}=\frac{M_{\text {gas }}}{M_{H_{2}}}=\frac{M_{\text {gas }}}{2} M_{\text {gas }}=2 V.D.$$

Mole-mole analysis :

PYQ-2024-Solutions-Q9 , PYQ-2024-Genral_Organic_Chemistry-Q9, PYQ-2023-Solutions-Q10

Concentration terms :

Molarity (M) :

PYQ-2024-Solutions-Q4, PYQ-2024-Solutions-Q5

$$ \therefore \quad \text { Molarity }(\mathrm{M})=\frac{\mathrm{W} \times 1000}{(\text { Mol. wt of solute }) \times \mathrm{V}_{\text {in ml }}} $$

Molality (m) :

PYQ-2024-Solutions-Q7

$$ Molality =\frac{\text{number of moles of solute} }{\text{mass of solvent in gram} }\times 1000 $$

$$ = \frac{1000 \times W_1}{M_1 \times w_2} $$

Mole fraction $(x)$ :

PYQ-2024-Solutions-Q10

$\therefore \quad$ Mole fraction of solution $\left(\mathrm{x}_{1}\right)=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{N}}$

$\therefore \quad$ Mole fraction of solvent $\left(\mathrm{x}_{2}\right)=\frac{\mathrm{N}}{\mathrm{n}+\mathrm{N}}$

$\quad \quad x_{1}+x_{2}=1$

% Calculation :

(i) %w/w = $\frac{\text { mass of solute in gm }}{\text { mass of solution in gm}} \times 100$

(ii) %w/v = $\frac{\text { mass of solute in gm}}{\text { Volume of solution in ml}} \times 100$

(iii) %v/v = $\frac{\text { Volume of solute in ml}}{\text { Volume of solution }} \times 100$

Conversions :

(1) Mole fraction of solute into molarity of solution $M=\frac{x_{2} \rho \times 1000}{x_{1} M_{1}+M_{2} x_{2}}$

(2) Molarity into mole fraction $ x_2 = \frac{MM_1 \times 1000}{\rho \times 1000 - MM_2} $

(3) Mole fraction into molality $m=\frac{x_{2} \times 1000}{x_{1} M_{1}}$

(4) Molality into mole fraction $ x_2 = \frac{mM_1}{1000 + mM_1} $

(5) Molality into molarity $M=\frac{m \rho \times 1000}{1000+\mathrm{mM}_{2}}$

(6) Molarity into Molality $\mathrm{m}=\frac{\mathrm{M} \times 1000}{1000 \rho-\mathrm{MM}_{2}}$

$M_{1}$ and $M_{2}$ are molar masses of solvent and solute. $\rho$ is density of solution $(\mathrm{gm} / \mathrm{mL})$

$M=$ Molarity (mole/lit.), $m=$ Molality (mole $/ \mathrm{kg}$ ), $x_{1}=$ Mole fraction of solvent, $x_{2}=$ Mole fraction of solute

Average/Mean atomic mass :

$$ A_{x}=\frac{a_{1} x_{1}+a_{2} x_{2}+\ldots . .+a_{n} x_{n}}{100} $$

Mean molar mass or molecular mass

$$ M_{\text {avg. }}=\frac{n_{1} M_{1}+n_{2} M_{2}+\ldots . . n_{n} M_{n}}{n_{1}+n_{2}+\ldots n_{n}} \quad \text { or }$$

$$\quad M_{\text {avg. }}=\frac{\sum_{j=1}^{j=n} n_{j} M_{j}}{\sum_{j=1}^{j=n} n_{j}} $$

Calculation of individual oxidation number :

Formula : Oxidation Number $=$ number of electrons in the valence shell - number of electrons left after bonding

Concept of Equivalent weight/Mass:

For elements, equivalent weight (E)= $ \frac{\text { Atomic weight }}{\text { Valency } \text { factor }}$

For acid/base, $ \quad \mathrm{E}=\frac{\mathrm{M}}{\text { Basicity / Acidity }}$

Where $\mathrm{M}=$ Molar mass

For O.A/R.A, $ \quad E=\frac{M}{\text { no. of moles of } \mathrm{e}^{-} \text {gained /lost }}$

Equivalent weight (E) =$\frac{\text { Atomic or molecular weight }}{v . f .}$

(v.f. $=$ valency factor)

Concept of number of equivalents

No. of equivalents of solute $=\frac{W t}{E q \cdot \text { wt. }}=\frac{W}{E}=\frac{W}{M / n}$

No. of equivalents of solute $=$ No. of moles of solute $\times$ v.f.

Normality (N) :

Normality $(N)=\frac{\text { Number of equivalents of solute }}{\text { Volume of solution (in litres) }}$

Normality $=$ Molarity $\times$ v.f.

Calculation of valency Factor :

$\mathrm{n}$ -factor of acid $=$ basicity $=$ no. of $\mathrm{H}^{+}$ ion(s) furnished per molecule of the acid.

$\mathrm{n}$ -factor of base $=$ acidity $=$ no. of $\mathrm{OH}^{-}$ ion(s) furnished by the base per molecule.

At equivalence point

$ \quad N_1 \ V_1 = N_2 \ V_2 $

$\quad n_{1} M_{1} V_{1}=n_{2} M_{2} V_{2}$

Volume strength of $ H_2 O_2 $ :

$20V H_2 O_2 $ means one litre of this sample of $H_2 O_2$ on decomposition gives 20 lt. of $O_2$ gas at S.T.P.

Normality of $H_2O_2$ (N) = $\frac{ { \text Volume, strength of H_2O_2}}{5.6}$

Molarity of $H_2O_2$ (N) = $\frac{ { \text Volume, strength of H_2O_2}}{11.2}$

Measurement of Hardness :

Hardness in ppm $=\frac{\text { mass of } \mathrm{CaCO}_{3}}{\text { Total mass of water }} \times 10^{6}$

Calculation of available chlorine from a sample of bleaching powder :

% of $Cl_2$ = $\frac{3.55 \times x \times V(mL)}{W(g)} $ where $\mathrm{x}=$ molarity of hypo solution

and v=m l of hypo solution used in titration.