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(a) How can you illustrate the law of multiple proportions by using oxides of metals containing 78.7% and 64.5% of the metal? (b) Match the following:1/12th the mass of C12 atom - 1 mole 1 g of hydrogen atom – amu 22.4 L O at NTP – gram mole 180 g of glucose – gram atom 6.022 × 1023 particles – molar volume |
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Answer» (a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively. First Oxide : Mass of oxygen = 100 – 78.7 = 21.3 g No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{78.7}{21.3}\) = 3.7g Second oxide: Mass of oxygen = 100 – 64.5 = 35.5 g No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}\) = 1.9g The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio. (b) 1/12th the mass C12 atom – amu 1 g of hydrogen atom – gram atom 22.4 L O2 at NTP – molar volume 180 g of glucose – gram mole 6.022 × 1023 particles – 1 mole |
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