Tiêu đề 1. Stoichiometry.pdf ✅

Stoichiometry 1 1 Stoichiometry QUICK LOOK The concentration of a solution can be expressed in number of ways. Normality It is defined as the number of equivalent of a solute present in 1 litre of solution.   Equivalent of a solute Eq N = Volume of solution in litre V in L  . . .(i)  Equivalent of substance (Eq) q Mass of substance E Equivalent mass of substance  Mass of solute 100 N Equivalent mass of solute V in mL    or N E V in mL   W 1000    . . .(ii) Equivalent =   Mass of solute N V in litre Equivalent mass of solute   . . .(iii) Also, Milli equivalent of solute = N V in mL    . . .(iv) Mass of solute Meq. of solute N 1000 Eq. mass of solute   . . .(v) Note  Equivalent and milli equivalent reacts in equal number to give the same number of equivalent of milli equivalent of products.  An equivalent represents the mass of material providing Avogadro’s number of electrons from reacting unit.  Normality of any solute depends up on the nature of reaction (like equivalent mass) but molarity not. Molarity: It is defined as the number of moles of solute present in 1 litre solution. Moles of a solute M Volume of solution in litre  Mass of solute 1000 M Molar mass of solute V (in mL)    Molarity Molar mass V in mL   W 1000 in mL    . . .(vi)    in litre Mass of solute Moles M V Moles mass of solute   . . . (vii)  Milli moles M V in mL   Mass of solute 1000 Molar mass of solute  . . . (viii) Note  Moles and milli moles of reactants react according to stoichiometry of equation and give products accordingly.    Moles Molarity = Volume in litre   Equivalent Normality = Volume in litre  M Moles W Eq. mass 1 ; N Equivalent Molar mass W Valence factor     Molarity × Valence factor = Normality . . .(ix)  Analytical molarity (usually molarity) and equilibrium molarity are two different terms. Equilibrium molarity represents the moles of particular species in one litre of solutions.  The equilibrium molarity of a strong electrolyte is zero. For example consider analytical molarity of 1 M HCl. The equilibrium molarity of HCl is zero because of 100% ionization as. + – HCl H +Cl 1 00 0 11  Whereas equilibrium molarities of + H and – Cl and 1 M each. Similarly a solution of a strong electrolyte say A Bx y having analytical molarity 1 M shows equlilibrium molarities of y AB,A x y  and x- B as 0 M, xM and yM respectively.  In case of weak electrolytes (either a weak acid or weak base) analytical molarity of C HA has the equilibrium molarities of HA, H  and A are C1 ,C     and C respectively. Since it shows partial dissociation + HA H + 1 0 1         A ( is degree of dissociation)  In case of non electrolytes analytical molarity and equilibrium molarity are same.
2 Quick Revision NCERT - CHEMISTRY Molality: It is defined as number of moles of solute present in 1 kg of solvent. Moles of solute m Mass of solute in kg  . . .(x) Mass of solute 1000 m Molar mass of solute Mass of solvent in gram    . . .(xi) Mass of solvent = mass of solution–mass of solute Note: Molarity, normality and molality are extensive properties. Strength of Solution: It is expressed as amount of solute (in g) in 1 litre solution.   Mass of solute w S Volume of solution in litre V in L   . . .(xii)    w S N E×V in L E   . . .(xiii) Also, S NE   or S M Molar mass   . . .(xiv) In Terms of Percentage % by mass (or w/w) Mass of solute × 100 Mass of solution  . . .( xv) % by strength (or V/V) Volume of solute × 100 Volume of solution  . . .( xvi) % by volume (or mass/vol) Mass of solute ×100 Volume of solution  . . .( xvii) A solution is 65% by... 65% by mass, i.e., 100 g solution has 65 g solute 65% by strength, i.e., 100 mL solution has 65 mL solute 65% by volume, i.e., 100 mL solution has 65 g solute Note  Use of Specific Gravity: Specific gravity term is used for density in case of solution. The term ‘density’ refers for the ratio (mass/volume) for pure solvent. Mass of solvent Density= Volume of solvent i.e., density is the mass of 1 mL of solvent.  Density is the mass per unit volume at the specified temperature, usually g/mL or g/cm3 at 20°C. Specific gravity is defined as the ratio of the mass of a solution at 20°C, to the mass of an equal volume of water at 4°C (or sometimes 20°C). Thus, specific gravity is a dimensionless quantity. The density of water at 4°C is 1.0000 g / mL . Density and specific gravity are equal when referred to water at 4°C. When specific gravity is referred to water at 20°C, then (since density of water at 20°C is 0.99823 g / mL) Density = specific gravity  0.99823 Mole Fraction Mole fraction of solute XA    A Moles of solute X Total moles present in solution  Moles of solute n Moles of solute + Moles of solvent n N    Mole fraction of solvent XB    B Moles of solvent X = Total moles present in solution Moles of solvent N = = Molesof solute + Moles of solvent n N Also,    A n w/m X n N w/m W/M     . . .(xviii)    B N W/M X n N w/m W/M     . . .( xix) Also, for a binary system XX1 A B   . . .( xx) i.e., sum of all the mole fractions in a solution is unity. Also, A B X n X N  . . .( xxi) Note  Molality, % by mass, mole fractions are independent of temperature since all these involve mass which does not depend upon temperature.  Standard Solution: The solution whose molarity or normality is known. A standard solution is prepared by dissolving an accurately weighed quantity of a highly pure material called a primary standard (e.g., oxalic acid, Borax 247 2 (Na B O 10H O)  N 23 2 a CO 10H O  and potassium acid phthalate, etc.) and diluting to an accurately known volume of solution. If the material is not sufficiently pure (e.g., NaOH) a solution is prepared to give approximately the desired concentration and this is standardized by titrating a weighed quantity of primary standard. Such solutions are called secondary standard.  Normal or Molar Solution: Solution having normality = 1 N or molarity = 1 M respectively.  On diluting a solution: Normality, molarity and molality change with dilution whereas, equivalent, mill equivalent, moles or milli moles of solute do not change or during dilution conservation of moles of solute take place.  The volume of a liquid increases with temperature. It is therefore, 5 litre of C H6 6 masses more in winter than in summer.

4 Quick Revision NCERT - CHEMISTRY MULTIPLE CHOICE QUESTIONS Significant Figures, Units for Measurement, Matter and Separation of Mixture 1. In the final answer of the expression 5 (29.2 20.2)(1.79 10 ). 1.37   The number of significant figures is: a. 1 b. 2 c. 3 d. 4 2. The prefix zepto stands for: a. 9 10 b. 12 10 c. 15 10 d. 21 10 3. The number of significant figures in 60.0001 is: a. 5 b. 6 c. 3 d. 2 Laws of Chemical Combination 4. Which of the following is the best example of law of conservation of mass? a. 12 g of carbon combines with 32 g of oxygen to form 44 g of CO2 b. When 12 g of carbon is heated in a vacuum there is no change in mass c. A sample of air increases in volume when heated at constant pressure but its mass remains unaltered d. The weight of a piece of platinum is the same before and after heating in air 5. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as, X Y R S.  The relation which can be established in the amounts of the reactants and the products will be a. nm pq   b. nm pq   c. n m d. p  q Atomic, Molecular and Equivalent Masses 6. Sulphur forms the chlorides S Cl and SCl . 22 2 The equivalent mass of sulphur in SCl2 is: a. 8 g/mole b. 16 g/mole c. 64.8 g/mole d. 32 g/mole 7. What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together? a. 0.1 M b. 0.2 M c. 0.05 M d. 0.25 M 8. 74.5 g of a metallic chloride contain 35.5 g of chlorine. The equivalent weight of the metal is: a. 19.5 b. 35.5 c. 39.0 d. 78.0 9. One litre of a gas at STP weight 1.16 g it can possible be: a. C2H2 b. CO c. O2 d. CH4 10. 1.25 g of a solid dibasic acid is completely neutralised by 25 ml of 0.25 molar Ba(OH)2 solutions. Molecular mass of the acid is: a. 100 b. 150 c. 120 d. 200 11. The atomic weights of two elements A and B are 40 and 80 respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B? a. y 2 b. y 4 c. y d. 2y 12. 1.520 g of the hydroxide of a metal on ignition gave 0.995 gm of oxide. The equivalent weight of metal is: a. 1.520 b. 0.995 c. 19.00 d. 9.00 13. What should be the equivalent weight of phosphorous acid, if P=31; O=16; H=1? a. 82 b. 41 c. 20.5 d. None of these 14. The weight of a molecule of the compound C H60 122 is: a. 21 1.4 10  g b. 21 1.09 10  g c. 23 5.025 10  g d. 23 16.023 10  g 15. Caffeine has a molecular weight of 194. If it contains 28.9% by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is: a. 4 b. 6 c. 2 d. 3 16. A gaseous mixture contains CH4 and C H2 6 in equimolecular proportion. The weight of 2.24 litres of this mixture at NTP is: a. 4.6 g b. 1.6 g c. 2.3 g d. 23 g 17. Volume of a gas at STP is 7 1.12 10  cc. Calculate the number of molecules in it: a. 20 3.01 10  b. 12 3.01 10  c. 23 3.01 10  d. 24 3.01 10  18. The number of moles of oxygen in 1 L of air containing 21% oxygen by volume, in standard conditions, is: a. 0.186 mol b. 0.21 mol c. 2.10 mol d. 0.0093 mol