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# chemistrymoles.doc

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Department
Psychology
Course Code
PSYCH-114
Professor
David Hartstein

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S 4 chemistry / revision exercise / mole concept / 03-04 / P.1 Mole concept 1. Find the number of moles of ions in (a) 2 moles of Fe 2SO )4 3 From the formula, we know that 1 mole of Fe 2SO 4 3ontains 2 moles of Fe 3+ 2− ions and 3 moles of SO4 ions. 3+ ∴ No. of moles of Fe ions = 2× 2 = 4 No. of moles of SO 42 ions = 3× 2 = 6 Total no. of moles of ions = 4 + 6 = 10 (b) 0.2 moles of Al(NO )3 3 ∴ No. of moles of Al ions = 0.2 No. of moles of NO 3ons = 0.2× 3 = 0.6 Total no. of moles of ions = 0.2 + 0.6 = 0.8 2. Given 1.6 g of methane (CH )4 find (a) number of moles of CH 4 Molar mass of CH 4 = 12 + 1 × 4 = 16 gmol −1 No. of moles of CH 4 = mass of CH /4olar mass of CH 4 = 1.6/16 = 0.10 (b) number of molecules of CH 4 No. of molecules of CH4= no. of moles× Avogadro Number 23 = 0.10 × 6.02 × 10 = 6.02 × 1022 (c) number of H atoms. One CH m4lecule contain 4 H atoms, ∴ no. of H atoms = 6.02 × 1022× 4 = 2.41 × 1023 3. Find the mass of (a) 1 H2O molecule (b) 1 Cu atom 23 23 18 / 6.02× 10 63.5 / 6.02× 10 −23 −22 3.0 × 10 g 1.05 × 10 g (c) 1 Na ion (d) 1 OH ion 23 23 23 / 6.02× 10 17 / 6.02 × 10 3.8 × 10−23g 2.8 × 10 23g (e) 1 neutron (f) 1 electron 23 23 1.0 / 6.02× 10 0.00055 / 6.02 × 10 1.7 × 10−24g 9.1 × 10 −28g 4. Find the mass of 22 24 23 6.02 × 10 lead atoms 3.01× 10 carbon dioxide 3.01 × 10 sulphate ions molecules No. of moles of Pb atoms No. of moles of CO 2 No. of moles of SO 4− ions = 6.02 × 10 / 6.02 × 1023 molecules = 3.01 × 10 / 6.02 × 10 23 24 23 = 0.100 = 3.01× 10 / 6.02 × 10 = 0.500 = 5.00 Molar mass of SO 42 S 4 chemistry / revision exercise / mole concept / 03-04 / P.2 Mass of Pb atoms Molar mass of CO = 32 + 16 × 4 = 96 g 2 = 0.100 × 207 = 12 + 16 × 2 = 44 g Mass of SO 4−ions = 20.7 g Mass of CO m2lecules = 0.500 × 96 = 5.00 × 44 = 48 g = 220 g 5. (a) How many molecules are there in 3.00 moles of oxygen molecules? Number of oxygen molecules 23 = 3.00× 6.02 × 10 = 1.806× 10 24 (b) How many ions are there in 0.600 moles of potassium ions? Number of potassium ions = 0.600× 6.02 × 1023 = 3.612× 10 23 (Relative atomic masses: O = 16.0, K = 39.0) 6. Calculate the number of moles of atoms in (a) 127 g of copper Number of moles of copper = 127 / 63.5 = 2 mol (b) 12.8 g of sulphur. Number of moles of sulphur = 12.8 / 32.0 = 0.4 mol (Relative atomic masses: Cu = 63.5, S = 32.0) 7. How many atoms are there in (a) 2.50 moles of oxygen atoms? Number of oxygen atoms 23 = 2.50× 6.02 × 10 = 1.505× 10 24 (b) 6.00 g of magnesium atoms? Number of mole of magnesium atoms = 6.00 / 24.0 = 0.25 mol Number of magnesium atoms 23 = 0.25× 6.02 × 10 = 1.505× 10 23 8. What is the mass of 2.50 moles of magnesium atoms? (Relative atomic mass: Mg = 24.0) Mass of magnesium atoms = 2.50 × 24.0 = 60.0 g S 4 chemistry / revision exercise / mole concept / 03-04 / P.3 − 9. Apure sample of calcium chloride CaCl w2s found to contain 7.10 g of Cl ions. + What mass of Ca ions does the sample contain? Number of moles of Cl ions= 7.10 / 35.5 = 0.200 2+ − The formula CaCl s2ows that the ratio of Ca ions to Cl ions is 1 : 2, hence no. of moles of Ca ions = 0.200 / 2 = 0.100 ∴ Mass of Ca ions in the sample = 0.100 × 40 = 4.0 g 10. Ametal M ionizes to give M ions. If atomic mass of M is 24, and 1.2 g of M ionize to give 6.02× 10 electrons, calculate n (the charge on each ion of M). No. of moles of electrons given = no. of electrons /Avogadro Number = 6.02 × 10 / 6.02 × 10 = 0.100 No. of moles of M ionized = 1.2 / 24 = 0.050 ∴ 0.050 mole of M gives 0.100 mole of electrons on ionization. Thus 1 mole of M gives 0.100 / 0.050 = 2 moles of electrons. Since the charge on an ion of M is numerically equal to the number of moles of electrons given by 1 mole of M, each ion of M carries 2 charges, i.e. n = 2. 11. Complete the table below: Substance Molar mass of Mass of Number of Number of substance substance moles of molecules/ present substance formula units present present −1 23 Sulphuric acid 98 g mol 58.8 g 0.6 mol 3.612 × 10 (H SO ) 2 4 −1 22 Sodium 40 g mol 2.0 g 0.05 mol 3.01× 10 hydroxide (NaOH) Potassium 138 g mol −1 331.2 g 2.4 mol 1.44× 1024 carbonate (K 2O )3 (Relative atomic masses: C = 12.0, H = 1.0, K = 39.0, Na = 23.0, O = 16.0, S = 32.0) 12. If we breathed in 3913 × 10 molecules of air pollutant, nitrogen dioxide (NO 2, how many grams of NO wou2d we breathe in? (Relative atomic masses: N = 14.0, O = 16.0) Number of mole of nitrogen dioxide = 3913 × 10 / 6.02 × 10 23 −3 = 6.5× 10 mol Mass of nitrogen dioxide = 6.5× 10−3× (14.0 +
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