b. number of moles of chlorine = 0.168 ÷ 24 = 0.007 mol
c. mass of chlorine gas in the sample = 0.007 × 71 = 0.5g
4a. number of moles in the gas = 2 ÷ 24 = 0.083
bi. 0.83 moles of the gas has a mass of 4g.
1 mole of the gas has a mass of 48.2g.
bii. relative molecular mass of the gas = 48.2
c. O3
5a. relative formula mass of NaCl = 23 + 35.5 = 58.5
b. moles of NaCl in 500cm3
= 0.212mol
c. number of NaCl units in 500cm3
= 1.27 × 10 23
d. concentration of the solution in mol/dm3 = 0.212/ 500 × 1000
= 0.424 mol/dm3
6a. relative formula mass of ethanol = 12 + 12 + 6 + 16 = 46
b. concentraion of the solution in g/dm3 = 12.5/200 × 1000
= 62.5g/dm3
c. concentration of the solution in mol/dm3 = 62.5/46
= 1.36 mol/dm3