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| Numericals Based on Mole Concept |
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 1 mole of an atom = 1 gram atomic weight of an atom |
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 1 mole of a molecule = 1 gram molecular weight of molecule |
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 1 mole of a gas = 22.4 liters of gas at STP |
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 1 mole of a substance = 6.023 x 1023, atoms, molecules or ions |
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 Molecular weight = 2 x Vapor density |
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 1 molar volume = 22.4 dm3 /L at STP |
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| Example: 1 |
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| Calculate the volume occupied by 2.8 g of N2 at STP. |
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| Solution |
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| Molecular weight of N2 = 2 x 14 = 28 g |
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| 28 g of N2 at STP occupies = 22.4L |
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 2.8 g of N2 at STP = ? |
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| 28 g of N2 = 22.4L |
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| 2.8g of N2 = ? |
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| 2.8 g of N2 at STP occupies a volume of 2.24L. |
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| Example: 2 |
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| Calculate gram molecular weights of the following gases: |
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| a. N2 (if 360 cm3 at STP weighs 0.45g) |
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| b. Cl2 (if 308 cm3 at STP weighs 0.97g) |
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| Solution: |
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| a. 360 cm3 of N2 = 0.45g |
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| 22.4L of gas = 1 gram molecular weight |
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| 22.4L = 22,400 cm3. (1L = 1000 cm3) |
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| 360 cm3 of N2 = 0.45g |
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| 22,400 cm3 of N2 = ? |
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| Gram molecular weight of N2 is 28 g. |
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| b. 308 cm3 Cl2 = 0.979 g |
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| 22.400 cm3 of Cl2 = ? |
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| Molecular weight of Cl2 = 71.9 g |
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| Example: 3 |
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| What is the volume of 32g of sulphur dioxide measured at STP? |
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| Solution |
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| Molecular formula = SO2 |
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| Molecular weight = 1 x 32 + 2 x 16 = 64g |
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| 64g of SO2 occupies 22.4L |
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| 32 g of SO2 = ? |
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| Volume of 32 g of SO2 is 11.2 liters. |
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| Example: 4 |
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| Calculate the volume at S.T.P. of 7.1g of chlorine. |
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| Solution: |
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| Cl = 35.5 |
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| 1 mole of a substance = 22.4L |
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| 1 Mole of a substance = 1 GMM |
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| 1 GMM of Cl2 = 71 g |
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| 71 g of Cl2 = 22.4 L |
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| 7.1 g of Cl2 = ? |
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| 7.1 g of Cl2 will occupy a volume of 2.24 liters. |
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| Example: 5 |
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| Calculate the number of moles of nitrogen in 7g of nitrogen. |
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| Solution: |
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| 1 mole of N2 = 1 GMM |
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| 1 mole of N2 = 2 x 14g = 28g |
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| 1 mole = 28 g |
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| ? = 7 g |
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| 7g of nitrogen is equal to 0.25 moles. |
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| Example: 6 |
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| Calculate the mass of 0.4 moles of water. |
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| Solution: |
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| 1 GMM of water (H2O) = 2 x 1 + 16 = 18g. |
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| 18g = 1 mole |
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| Xg = 0.4 moles |
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| 0.4 moles of water weighs 7.2g. |
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| Example:7 |
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| Calculate the vapor density and molecular weight of CO2 gas occupying 200 mL and weighing 0.40 g at STP. |
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| Solution: |
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| 200 mL of CO2 = 0.40 g |
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| 22.4 L of CO2 = 12 + 32 (44g) |
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| Molecular weight = 2 x vapor density |
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| = 2 x 22.22 = 44.44 |
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| Note |
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| Molecular weight has no units. |
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| Example: 8 |
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| Calculate the gram atoms present in 8g of oxygen. |
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| Solution |
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| Example: 9 |
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| Calculate the gram molecules present in 45g of water. |
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| Solution |
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| 1 molecule = 2 + 16g = 18g of H2O |
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| ? = 45 g of H2O |
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| Example: 10 |
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| Calculate the number of molecules in 500g of sodium chloride. |
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| Solution: |
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| 1 GMM = 6.023 x 1023 molecules |
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| 23 + 35.5 g = 58.5g = 1 GMM of NaCl |
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| 58.5g = 6.023 x 1023 molecules |
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| 500g = ? |
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| The number of molecules in 500g of sodium chloride is |
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| 34.2 x 6.023 x23 molecules |
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| Example: 11 |
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| 0.48 g of a gas forms 100 cm3 of vapors at STP. Calculate the gram molecular weight of the gas. |
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| Solution: |
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| 22.4L of a gas = 1 GMM |
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| 100 cm3 of gas = 0.48 g |
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| 22.4 x 1000 of gas = ? |
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| Gram molecular weight of the gas is 107.52 g. |
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| All elements are represented by symbols and all compounds represented by chemical formulae indicating the number of atoms of elements and also the proportion of the atoms in the compound. |
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| Example |
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| Hydrogen atom is represented as H. |
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| Hydrogen molecule is represented as H2. |
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| A compound of hydrogen, water is represented as H2O. |
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| In H2O - proportion of atoms H : O = 2 : 1 |
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| Knowing the proportion of atoms in a compound, the percentage composition can be calculated. |
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| Percentage composition of a compound is the percent by weight of each element present in it. |
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Percentage composition of an element  |
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| Example:1 |
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| Calculate the percentage by weight of all the elements present in calcium carbonate. |
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| Solution: |
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| Calcium carbonate = CaCO3 Ca = 40, C = 12, O = 16 |
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| GMM = 1 x 40 + 1 x 12 + 3 x 16 |
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| = 40 + 12 + 48 = 100 g |
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| Example: 2 |
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| Calculate the percentage by weight of potassium in potassium dichromate. |
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| Solution |
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| Potassium dichromate = K2Cr2 O7 |
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| GMM = (2 x 39) + (2 x 52) + (7 x 16) |
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| = 78 + 104 + 112 g = 294 g |
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