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Deptt . Of Applied Sciences Govt. Polytechnic College For Girls Patiala Presented By-

Deptt . Of Applied Sciences Govt. Polytechnic College For Girls Patiala Presented By- Dr. Raman Rani Mittal M.Sc., M.Phil , Ph.D. (Chemistry). SOLUTIONS & MOLE CONCEPT. Contents. Mole Concept Atomic & Gram atomic mass Molecular mass & Gram molecular mass

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Deptt . Of Applied Sciences Govt. Polytechnic College For Girls Patiala Presented By-

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  1. Deptt. Of Applied Sciences Govt. Polytechnic College For Girls Patiala Presented By- Dr. Raman RaniMittal M.Sc., M.Phil, Ph.D. (Chemistry)

  2. SOLUTIONS&MOLE CONCEPT

  3. Contents • Mole Concept • Atomic & Gram atomic mass • Molecular mass & Gram molecular mass • Mole concept & its importance • Solutions • Methods of expressing concentrations of solutions

  4. Mole Concept

  5. Atomic & Molecular Mass • Matter is made up of atoms & molecules • Mass of matter is due to atoms & molecules • Mass of an atom is called as atomic mass & mass of molecule is termed as molecular mass • The actual mass of an individual atom or molecule is extremely small. This mass can not be expressed in grams. • To express the masses of atoms & molecules, the unit called Atomic Mass Unit (a.m.u.) is introduced.

  6. The atomic mass unit (amu) may be defined as one-twelfth of the actual mass of an atom of carbon (Carbon-12 isotope). • The atomic mass of an element tell us as to how many times an atom of the element is heavier than 1/12th of an atom of Carbon (C-12) Mass of an atom of an element 1/12 x mass of an atom of carbon(C-12) Atomic mass =

  7. For example, atomic mass of oxygen is 16 a.m.u. It means that an atom of oxygen is 16 times heavier than 1/12th of mass of carbon atom (C-12) The atomic mass of an element is the average relative mass of its atoms as compared with an atom of carbon taken as 12 a.m.u.

  8. Gram Atomic Mass Gram atomic mass It is the quantity of an element whose mass in grams is numerically equal to its atomic mass. In simple words, atomic mass of an element expressed in grams is the gram atomic mass or it is also called gram atom For example, the atomic mass of oxygen is 16 a.m.u. and thus gram atomic mass of oxygen is 16 g. Mass in grams Gram atomic mass Number of gram-atom

  9. Molecular mass Molecular mass of a substance ( element or compound) may be defined as the average relative mass of a molecule of the substance as compared with mass of an atom of carbon (C12)taken as 12 a.m.u.

  10. In other words, Molecular mass expresses as to how many times a molecule of the substance is heavier than 1/12th of the mass of an atom of carbon. For example, A molecule of CO2 is 44 times heavier than 1/12th of the mass of carbon atom. Hence, molecular mass of CO2 is 44 a.m.u.

  11. The molecular mass is obtained by adding together the atomic masses of various atoms present in a molecule. For example, The molecular formula of carbon dioxide is CO2. Hence , its molecular mass is = Atomic mass of carbon + 2 x (Atomic mass of oxygen) = 12 + 2 x 16 = 44 a.m.u.

  12. Gram Molecular Mass Gram molecular mass is the quantity of a substance whose mass in grams is numerically equal to its molecular mass. In other words, molecular mass of a substance expressed in grams is called gram molecular mass. It is also known as gram molecule. For example, the molecular mass of oxygen is 32 and therefore, its gram molecular mass is 32 g. Mass in grams Gram molecular mass No. of gram molecule

  13. Example : Calculate the molecular mass of calcium carbonate. Solution : Molecular mass of calcium carbonate (CaCO3) = 1 x atomic mass of Ca + atomic mass of C + 3 x atomic mass of O = 1 x 40 + 12 + 3 x 16 = 40 + 12 + 48 = 100 amu

  14. Mole Concept • In chemistry we use a unit mole to count particles (atoms, ions or molecules). • A mole is a collection of 6.023 x 1023 particles irrespective of their nature. • The no. 6.023 x 1023 is called Avogadro’snumber and is denoted by No. 1 Mole = 6.023 x 1023 particles

  15. For example, 1 mole of atoms = 6.023 x 1023atoms 1 mole of molecules = 6.023 x 1023molecules 1 mole of ions = 6.023 x 1023ions 1 mole of electrons = 6.023 x 1023electrons 1 mole of protons = 6.023 x 1023protons While using the term mole, it is important to indicate the kind of particles involved. 1 mole of H atoms = 6.023 x 1023atoms of H 1 mole of H molecules = 6.023 x 1023molecules of Hydrogen The mole is also related to the mass of the substance

  16. 1. Mole & Gram Atomic Mass • Mass of one mole of atoms of any element in grams is equal to its gram atomic mass or one gram atom. One mole of atoms = 6.023 x 1023atoms = Gram atomic mass of the element For example, The mass of 6.023 x 1023 atoms of oxygen is 16 grams.

  17. 2. Mole & Gram Molecular Mass Mass of one mole of molecules of any substance in grams is equal to its gram molecular mass. One mole of molecules = 6.023 x 1023 molecules = Gram molecular mass For example, • The mass of 6.023 x 1023molecules of sulphur dioxide (1 mole) is equal to 64 grams. • Similarly, the mass of 6.023 x 1023molecules of carbon dioxide (CO2) is equal to 44 grams.

  18. 3. Mole concept for Ionic compounds Mass of one mole of formula units of any ionic compound in grams is equal to its gram formula mass. One mole of formula units = 6.023 x 1023 formula units = Gram formula mass For example, The mass of 6.023 x 1023 formula units of NaCl or 6.023 x 1023 Na+ ions and 6.023 x 1023Cl- ions (one mole of NaCl) is equal to 58.5 g or 1 formula mass of NaCl.

  19. 4.Mole in terms of Volume One mole of any gas at S.T.P. (0oC and 760 mm pressure) occupies 22.4 litres. This volume is known as molar volume. For example, 1 mole (2 grams) of H2 gas = 22.4 litres at S.T.P. 1 mole (28 grams) of N2 gas = 22.4 litres at S.T.P. 1 mole (32 grams) of O2 gas = 22.4 litres at S.T.P. 1 mole (44 grams) of CO2 gas = 22.4 litres at S.T.P.

  20. One mole of a gas = 22.4 litres at S.T.P. =6.023 x 1023 molecules = Gram molecular mass For example, 1 mole H2 gas = 22.4 litresat S.T.P. = 6.023 x 1023molecules of H2 = 2g

  21. To calculate number of moles • For Elements No. of atoms of the element 6.023 x 1023 Mass of the element in grams Gram atomic mass Number of moles

  22. Example: Calculate the number of moles in 22 g of CO2. Solution: Molecular mass of CO2 = 12+2x16 = 44 amu Gram molecular mass of CO2 = 44 g Since, 44 g of CO2 make = 1 mole 18 6.023 x 1023g = 0.5 mole .∙. 22 g of CO2 will make

  23. (b) For Compounds No. of molecules of the compound 6.023 x 1023 Mass of the compound in grams Gram molecular mass No. of moles

  24. (c) For an assembly of things Number of things 6.023 x 1023 (d) 1 mole = gram molecular mass =6.023 x 10 23 = 22.4 L at S.T.P. Number of moles

  25. Importance of Mole concept • To calculate mass of one atom of element Gram atomic mass 6.023 x 10 23 Mass of an atom of element =

  26. Example: What is the mass of single atom of Hydrogen ? Solution : Atomic mass of Hydrogen = 1.008 amu Gram atomic mass of hydrogen = 1.008 g Thus 6.023 x 1023atomsof hydrogen have mass = 1.008 g 1.008 6.023 x 1023 =1.66 x 10- 24g Hence, a single atom of H will have mass =

  27. Example: What is the mass of single atom of Carbon? Solution : Atomic mass of Carbon =12 amu Gram atomic mass of carbon = 12 g Thus6.023 x 1023atomsof Carbon have mass = 12 g 12 6.023 x 1023 =1.99 x 10- 23g Hence, a single carbon atom will have mass =

  28. 2. To calculate mass of one molecule of the substance. Gram molecular mass 6.023 x 10 23 Mass of one molecule of substance

  29. Example:What is the mass of a single molecule of hydrogen? Solution: Molecular mass of H2 = 2 amu Gram molecular mass of H2 = 2 g Thus, 6.023 x10 23 molecules of H2 have mass = 2g 2 6.023 x10 23 =3.3 x 10 - 24 g .∙. A single molecule of H2 will have mass =

  30. Example:Calculate the mass of a single molecule of water? Solution: Molecular mass of H2O = 2x1 + 16 =18 amu Gram molecular mass of H2O = 18 g Thus, 6.023 x10 23 molecules of H2O have mass = 18g 18 6.023x1023= 2.99 x 10-23g .∙. A single molecule of H2O will have mass =

  31. 3. To calculate the number of atoms in a given mass of the element. No. of atoms in a given mass of element Mass of element in grams Gram atomic mass x 6.023x1023

  32. Example:Calculate the number of atoms in 20g of calcium (At. Mass of Ca = 40). Solution: Atomic mass of Ca = 40 amu Gram atomic mass of Ca = 40 g Thus 40 g of Ca contains = 6.023 x1023 atoms 6.023 x1023 40 = 3.01 x 1023atoms .∙. 20g of Ca will contain x20

  33. 4. To calculate the number of molecules in a given mass of substance. No. of molecules in a given mass of substance Mass of substance in grams Gram molecular mass X 6.023x1023

  34. Example: Calculate the number of molecules of methanein 0.80 g of methane. Solution:Molecular mass of methane (CH4) = 1x12 + 4x1 = 16 amu Gram molecular mass of CH4= 16 g Thus16gof CH4 contain =6.023 x1023molecules 6.023 x1023 16 = 3.01x1022 molecules .∙. 0.80g of CH4 will contain x 0.80

  35. 5. To calculate the volume occupied by a given mass of the gas at S.T.P. Volume occupied by a given mass of a gas at S.T.P. Mass of gas in grams Gram molecular mass of gas x22.4

  36. Example: Calculate the volume occupied by 16 g of oxygen at S.T.P. Solution: Molecular mass of O2 = 32 amu Gram molecular mass of O2 = 32 g Now 32 g of O2 at S.T.P. occupy volume = 22.4 L .∙. 16g of O2 at S.T.P. would occupy volume 22.4 32 = 11.2 litres = x16

  37. SOLUTIONS

  38. Solutions • Solution is a homogeneous mixture of two or more pure substances • Composition of solution can be varied within certain limits. • Solution results by dissolving a solute in a solvent. Solute + SolventSolution

  39. The substance which is dissolved & is present in lesser quantity is called solute. • The substance in which solute is dissolved & is present in greater quantity is called solvent. For example, 2g of sugar is dissolved into 50 ml of water to form a solution. In this case sugar is solute & water is solvent.

  40. Methods of expressing Concentration of a solution The concentration of solution is defined as the amount of solute present in the given quantity of the solution. It can be expressed in the following ways: 1. Strength 2. Molarity 3. Molality 4. Mole fraction 5. Normality 6. Mass percentage 7. Volume percentage

  41. Strength: It is the amount of solute in grams dissolved per litre of the solution. Thus, Mass of solute in grams Volume of solution in litres If ‘a’ grams of solute is dissolved in V ml of a given solution, then a x 1000 V It is expressed in g/L. Strength of solution = Strength =

  42. 2. Molarity (M): The number of moles of solute dissolved per litre of the solution is called molarity. Number of moles of solute Volume of solution in litres It is convenient to express volume in ml. So that Number of moles of solute Volume of solution in ml Molarity = x1000 M =

  43. And number of moles of solute can be calculated from the given mass of solute which is dissolved Given mass Gram molecular mass Thus, Given mass x 1000 Gram mol. mass x V Molarity changes with temperature. Moles of solute = M =

  44. Example:Calculate the molarity of the solution containing 0.5g of NaOH dissolved in 500 ml Solution: Given mass of NaOH = 0.5g Mol.mass of NaOH = 23 + 16 + 1= 40 0.5 40 Volume of solution, V = 500ml No. of moles of NaOHx 1000 V 0.0125 x 1000 500 No. of moles of NaOH = = 0.0125 Now, Molarity= = = 0.025 M

  45. Molality (m) = 3. Molality: The number of moles of solute dissolved per kg of the solvent is called molality. No. of moles of solute Mass of solvent in kg No. of moles of solute Mass of solvent in grams If ‘a’ grams of the soute is dissolved in W gram of the solvent, then a 1000 mol. Mass of solute W Molality does not changes with temperature. = x1000 m = x

  46. Example:Calculate the molality of an aqueous solution containing 4g of urea (Mol.mass = 60) in 500 g of water. Solution: Given mass of urea = 4g Mol. mass of urea = 60 Mass of water , W = 500g Given mass 1000 Mol. mass W 4 1000 60 500 = 0.133 m Since, Molality (m) = x = x

  47. 4. Mole fraction (x):It is the ratio of the number of moles of a component to the total number of moles of all the components (solute & solvent) in the solution. Suppose a solution contains nA moles of solute (A) and nB moles of solvent (B). Then, nA nA + nB nBnA + nB Mole fraction of solute (xA) = Mole fraction of solvent (xB) =

  48. The sum of mole fractions of all the components is always equal to one. nAnB nA + nBnA + nB Mole fraction being a ratio, is dimensionless property. xA + xB = + = 1

  49. Example:A solution containing 23g of ethanol & 90g of water. What is the mole fraction of ethanol & water in solution? Solution: Mass of ethanol (C2H5OH) = 23g Molecular mass of ethanol = 2x12 + 1x6 + 16 = 46 23 46 Mass of water = 90g Molecular mass of water = 18 .∙. Number of moles of ethanol = = 0.5

  50. .∙. Number of moles of water = 90 18 = 5 .∙. Total number of moles = 0.5 + 5 = 5.5 0.5 5.5 = 0.09 Mole fraction of water = 1 – 0.09 = 0.91 .∙. Mole fraction of ethanol =

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