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Saturday, April 22, 2023

Chemistry Class 12 All Formulas PDF Download

As a science student, you always look for chemistry formulas for class 12 pdf to download and use offline. After reading this article, chemistry class 12 all formulas pdf can be downloaded.



Chemistry Class 12 All Formulas PDF Download



Class 12 Chapter 1 Solutions Chemistry Formulas List



1). Mass percentage of solute= `\frac{W_B}{W_B+ W_A} ×100`


`W_B`=mass of solute 
`W_A`=mass of solvent 




2).Volume percentage of solute = `\frac{V_B}{V_B+ V_A} ×100`


`V_B`=volume of solute 
`V_A`=volume of solvent 




3). Mass-Volume percentage of solute = `\frac{W_B}{V_B+ V_A} ×100`


`V_B`=volume of solute 
`V_A`=volume of solvent 
`W_B`=mass of solute 




4). Parts per million(ppm)= `\frac{mass   of   solute}{Mass   of solution}×10^6`






5).Molarity (M)=`\frac{(\text{mass of solute in grams})}{(\text{molar mass of solute})} \times \frac{1}{(\text{volume of solution in litre})}`







6).Molarity (M) = `\frac{\text{Gram mole of solute}}{\text{Volume of solution in litres}}`





7).Molarity (M) = `\frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{volume of solution in mL}}`





8).Normality (N)= `\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in litres}}`





9).Normality (N) =` \frac{\text{mass of solute in grams}}{\text{Gram eq. mass of solute}} \times \frac{1}{\text{volume of solution in litres}}`






10).Normality (N) = `\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in litres}} \times \frac{1000}{\text{Volume of solution in mL or cm}^3}`






11).Molality (m) = `\frac{\text{Gram moles of solute}}{\text{Weight of solvent in Kg}}`






12).Molality (m) = `\frac{\text{mass of solute in grams}}{\text{molar mass of solute}} \times \frac{1}{\text{weight of solvent in Kg}}`







13).Molality (m) = `\frac{\text{mass of solutes in grams}}{\text{molar mass of solute}} \times \frac{1000}{\text{weight of solvent in grams}}`





14).Mole fraction of solute (`X_A`)= `\frac{n_A}{n_A + n_B}`


 



15).Mole fraction of solute (`X_B`)= `\frac{n_B}{n_A + n_B}`




16).   Solubility of a Gas in a Liquid (Henry’s law)

            P= KHX            [KH =  Henry’s law constant,  X= mole fraction of gas, P= partial pressure]




17).   Raoult’s law [Partial vapour pressure of two components of solution]


    `P_A= P_A^o X_A`

    `P_B= P_B^o X_B`



Where

 [ `P_A`  & `P_B`   are partial pressure of components A &B

   `P_A^o` & `P_B^o`   are vapour pressure in a pure state, `X_A`  & ` X_B`   mole fractions of A and B in solution.]






18). Lowering of Vapour Pressure

 

`X_B = \frac{P_A^0 - P_A}{P_A^0}`

 

`X_B = \frac{n_B}{n_A + n_B}`

 

`X_B = \frac{W_B/M_B}{W_A/M_A + W_B/M_B} = \dfrac{W_B M_A}{W_A M_B + W_B M_A}`



Where

[`X_B` = mole fraction of solute,

 

`n_A` and `n_B` are the number of moles of solvent and solute respectively,

 

`W_B` and `W_A`are the masses of solute and solvent,

 

`M_A` and `M_B` are the molecular masses of solvent and solute.]




19).   Elevation of boiling point

    `∆T_b=K_b m`


 Where 

[`∆T_b`=elevation of boiling point,`K_b`=molal elevation constant,m=molality of solution]




20).   Depression of Freezing point

            `∆T_f= K_f m`


  Where 

   

`∆T_f`= depression in freezing point,

`K_f` = molal elevation constant,

m =molality of solution.




21).   Calculation of molecular mass of an unknown non-volatile compound from depression of freezing point


 `M_B=\frac{K_f}{∆T_f} × \frac{W_B}{W_A}   ×1000`


Where

     MB= molar mass of solute

     WA=  mass of solvent

     WB=  mass of solute




22).   Osmotic pressure 

    `Ï€=\frac{n}{V} RT       or   `Ï€=CRT`

  Where     

 `Ï€=osmotic     pressure,C=molar concentration of solution,R=gas constant`




23).   Calculation of molar mass from osmotic pressure


`M_B=\frac{W_B RT}{Ï€V}`

  

Where

 MB= molecular weight of solute

WB=  mass of solute




24).  Van’t Hoff’s factor


`i = \frac{\text{normal molar mass}}{\text{abnormal molar mass}}`

 

 After reading this, the class 12 chemistry solutions formulas pdf can be downloaded from the given link below


 

`i = \frac{\text{observed colligative property}}{\text{theoretical colligative property}}`

 

 

`i = \frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}}`


Where 

i= Van't Hoff's factor



Here you have been given  all formulas of chemistry class 12 chapter-wise pdf 




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All Formula of Solid State Class 12




1.`\text{Packing efficiency} = \frac{\text{volume occupied by atoms}}{\text{volume of unit cell}}\times100`


i.`\text{Packing efficiency in hcp and ccp}  = \frac{\text{volume  of  4 speheres}}{\text{Total volume  of  unit  cell}}\times100`



ii.`\text{Packing efficiency in bcc}= \frac{\text{volume of  2  speheres}}{\text{Total volume of unit cell}} \times100`



iii. `\text{Packing efficiency in simple cubic cell}= \frac{\text{volume of 1  speheres}}{\text{Total volume of unit cell}} \times100`




2. Calculations involving unit cell dimensions


i. Mass of unit cell = `z\times m`  


 [z= number of atoms, m=mass of atom]


ii.Mass of atom(m)= `\frac{M}{N_A}`  


  [M= molar mass, NA=  Avogadro number]


iii.Density of unit cell(d)= `\frac{n\ \times\ M}{a^3\ \ \ \ \times\ \ N_A}`        


[n= number of atoms, M=molar mass, a=edge length of unit cell, NA=  Avogadro number]


3. Relation between distance (d) and the edge(a) of the unit cell


i. For simple cell   d = a


ii.For fcc       `d=\ \frac{a}{\sqrt2}` 


iii.For bcc       `d=\ \frac{\sqrt3a}{2}`


4. Relation between atomic radius (r) and the edge (a) of the unit cell


i.For simple cell  `r=\frac{a}{2}`


ii.For fcc    ` r=\frac{a}{2\sqrt2}`


iii.For bcc     `r=\frac{\sqrt3a}{4}`


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