SOLID STATE :- CRYSTAL LATTICE

A crystal is a regular arrangement of atoms, ions, or molecules, and is conceptually built up by the continuing translational repetition of some structural pattern. This pattern, or unit cell, may contain one or more molecules or a complex assembly of molecules. In three dimensions (3D), the unit cell is defined by three edge lengths a,b,c and three interaxial angles a, b, g. The different 3D crystal systems (triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal and cubic) arise from the seven fundamental unit cell shape

SOLID STATE , CRYSTAL LATTICE "BRAVAIS LATTICE"

PERIODIC TABLE

SOLUTION AND COLLIGATIVE PROPERTIES NUMERICALS FOR 12th std-:

“Solution and colligative properties” numericals for 12^th std-:

• 1.    Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.
• 2.    Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.        
• 3.    Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.
• 4. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
• 5.    Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
• 6.    Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
• 7.    Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL.
• 8.    The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?
• 9.  9.18 g of glucose, C₆H₁₂O₆, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? K_b for water is 0.52 K kg mol^-1.
• 10.  The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b for benzene is 2.53 K kg mol^–1
• 11.         11.45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate    (a) The freezing point depression and (b) The freezing point of the  solution.
• 12.           1.0 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^–1. Find the molar mass of the solute.
• 13.           200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein.
• 14.           Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
• 15.           Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
• 16.           2 g of benzoic acid (C₆H₅COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^–1. What is the percentage association of acid, if it forms dimer in solution?

PERIODIC TABLE

" APPLICATION OF POAC ( PRINCIPAL OF ATOMIC CONSERVATION) AND LAW OF CHEMICAL EQUIVALENCE IN NUMERICALS OF PHYSICAL CHEMISTRY OF JEE / ISEET.

1.    Mole concept

2.    POAC

3.    Equivalent method ( law of chemical equivalence)

POAC & equivalent method does not depend on stoichiometric coefficient, but one necessary condition is that we must have given reactant from which products are to be produced. i.e. balancing of reaction is not necessary for applying POAC and eq. method. POAC is used for simpler reaction.

For saving time Eq. Method should apply for solving complex/ redox reaction rather than POAC method.

POAC method:- principle of atomic conservation

Through the chemical reaction, total no. of atoms of reactant must be equal to products. If atoms are conserved, moles of those atoms shall also be conserved. Examples:-

NaNO₃     →   NaNO₂ + O₂

Reactant & product contains only 1 Na atom. So we can write as

1 * mole of NaNO₃ = 1 * mole of NaNO₂

1*[ wt. Of NaNO₃/ m.wt of NaNO₃] = 1*[ wt. Of NaNO₂/m.wt. of NaNO₂]

Similarly we can apply POAC on N & O atoms

1* mole of NaNO₃    = 1* mole of NaNO₂   (for N atoms)

3* mole of NaNO₃    = 2* mole of NaNO₂ + 2 * mole of O₂   (for O atoms)

Since one mole of NaNO₃, NaNO₂ & O₂ contain 3, 2, and 2 moles of O atoms respectively.

KClO₃ → KCl + O₂

For   K atom

1* mole of KClO₃ = 1* mole of KCl

3 * mole of KClO₃ = 2 * mole of O₂

Q.1 calculates the no. of O & P atoms in 1420 gm of P₄O₁₀. M.wt of P₄O₁₀  = 568

Mole of P₄O₁₀ = 1420/ 568 = 5 moles

1* mole of P₄O₁₀ = 4* mole of P

5 moles of   P₄O₁₀ = 20 moles of P

20 moles of P   =   20 * 6.022* 10 ²³ atoms of P = 12.044 * 10 ²⁴ atoms of P

Similarly O atom

1* mole of P₄O₁₀ = 10* mole of O

5 mole of P₄O₁₀ = 50 mole of O

50 mole of O = 50* 6.022 * 10²³ atoms of O

= 30.11 * 10²⁴ atoms of O

Equivalent method:-

Sum of equivalents( Eqs.) or milliequivalents ( meqs.) of all oxidising agents is equal to sum of Eqs. Or  meqs. Of all reducing agents.

i.e. sum of equivalents of oxidising agents = sum of equivalents of reducing agents.

Example{ 1} Suppose we have a mixture of  FeO+ Fe₃O₄ ,in this mixture we add acidic  KMnO₄.

Here Fe is in oxidation state of +2 in FeO and +8/3 in Fe₃O₄ i.e. less than the maximum Oxidation state of Fe (less than +3). Hence in the presence of oxidising agent, it has a tendency to oxidise itself. In KMnO_4,   Mn has +7 oxidation state {max. Oxd. State} so it can not be further oxidise. It can be only reduce to itself in the presence of any reducing agent so 

Meq. of FeO + meq. of Fe₃O₄ = meq. Of KMnO₄

Or n₁ * mole of Fe₃O₄ + n₂ * mole of FeO = n₃ * mole of KMnO₄

Or           N₁V₁       +     N₂V₁      =      N₃V₂

Example 2. If a mixture contains FeO+ Fe₃O₄ + Fe₂O₃ (which is not reacted to each other) and we add KMnO₄  in it initially, which oxidise Fe of mixture to completely to its higher oxidation state ( +3) then we write

Meq. Of FeO+ meq. of Fe₃O₄ = meq. Of KMnO₄

Because Fe is in max. Oxidation state(+3)  in  Fe₂O₃. So it can’t be further oxidise. so Fe₂O₃ does not react with  KMnO₄ .

If after this oxidation process we add excess of KI (reducing agent), it will reduce Fe₂O₃ present in the initial solution. In this case Fe₂O₃  is react as oxidising agent because of max. Oxd. State (+3) and  KI as reducing agent

How to prepare for ISEET 2014:- “Learn the fundamental concepts of Physics, Chemistry and Mathematics” what is the meaning of learning the concepts.

How to prepare for ISEET 2014

“Learn the fundamental concepts of Physics, Chemistry and Mathematics”: - everybody says that “learn fundamental concepts to crack any competitive exams, learn basic concepts. Don’t just remember, or mugup the things.” but do u know what is exactly meaning of “learning the fundamentals and basics” Let’s see what is the meaning of learning the concepts.

For example if you are learning about equivalent no., equivalent wt. Normally most students think that always equivalent wt. Of H₂SO₄ is 49. If you are thinking the same, means your concepts also not clear. It is not always 49.

Let’s clear the concept

 Equivalent wt = molecular wt. / n factor

Now n factor for acids = basicity of acids (no. Of total replaceable H⁺ ion)

Because there are two H in H₂SO₄ are replaceable. Its n factor is 2.

So eq. Wt = 98/2 =49

But it is always not true. n factor also depend on chemical reaction. If in any reaction H₂SO₄ only donate 1 H⁺ ion and gets converted into HSO₄^- for example :-

NaOH + H₂SO₄ →NaHSO₄ + H₂O

In the above reaction H₂SO₄  only donating 1 H⁺. So here n factor for H₂SO₄ is one 1. So eq.wt = 98/1= 98 in this reaction

But in the following reaction

2 NaOH + H₂SO₄ →Na₂SO₄ + 2H₂O

H₂SO₄ donate its all 2 H⁺ ion and gets converted into SO₄^-2 ion. So in this reaction n factor for H₂SO₄ is 2. So eq. Wt of H₂SO₄ in this reaction = 98/2 = 49.

Because H₂SO₄ have total 2 H⁺ to donate. So generally we assume that it is donating its all two H⁺. Hence generally we consider n factor for H₂SO₄ is two. But if in the question there is any reaction of H₂SO₄ is given then carefully see the reaction whether it is donating 1 or two H⁺. According to that your ans. of eq.wt. of H₂SO₄ would be 98 or 49. But if there is no such reaction of H₂SO₄ is given then we assume that H₂SO₄  is donating its all two H⁺ ion so n factor is assumed two and eq.wt is 49.

VOLUMETRIC CALCULATIONS:- NUMERICALS FOR IIT JEE / ISEET

VVVOLUMETRIC CALCULATION :- TRY TO SOLVE FOLLOWING NUMERICALS FOR IIT JEE/ ISEET

1.  A solution contains 1.2046* 10²⁴  HCl molecule in one dm³ of the solution. What is the normality of the solution?
2.   A solution containing 0.275 gm of NaOH required 35.4 ml of HCl for neutralization. What was the normality of the HCl?
3.   The formula mass of an acid is 82.0. In a titration, 100 cm³ of a solution of this acid containing 39.0 gm of this acid per litre were completely neutralized by 95.0 c.c. of NaOH per litre. What is the basicity of the acid?
4.  A solution contains Na₂CO₃ and NaHCO₃. 10 ml. of the solution required 2.5 ml of 0.1 M H₂SO₄ for neutralization using phenolphthalein as indicator. Methyl orange is than added when a further 2.5 ml of 0.2 M H₂SO₄ was required. Calculate the amount of   Na₂CO₃ in one litre of the solution.
5.    500 ml of a solution contains 2.65 gm of Na₂CO₃ and 4 gm of caustic soda; 20 ml. of this solution have been titrated each time against N/ 10 H₂SO_4.  Find the titer values if (i) methyl orange is taken as indicator, and also if (ii) phenolphthalein is taken as an indicator.
6.  25 ml. of a solution containing NaOH and Na₂CO₃ when titrated against N/2 HCl using phenolphthalein as indicator required 40 ml. of HCl. The same vol. of mix. When titrated against N/2 HCl using methyl orange as indicator required 45 cc of HCl. Calculate the amounts of NaOH and Na₂CO₃ in 1 litre of the mixture.
7.     Acidified KMnO₄   oxidizes oxalic acid to CO_2. what vol. in the litres of 10^-4 M KMnO₄ required to completely oxidize 0.5 litre of 10^-2 M oxalic acid in acid medium?                                                   

MOLECONCEPT AND LAW OF CHEMICAL EQUIVALENCE

If you are preparing for ISEET 2014 or  JEE OR NEET 2014 or any other medical or engineering competitive exam u have to start your preparation from physical chemistry "MOLE CONCEPT and LAW OF CHEMICAL EQUIVALENCE "  because its basic concept and also used in a no. of other topic also like in Faraday's law (electro chemistry), chemical and ionic equilibrium etc.