(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