(i)Gram atomic mass of Au=197g
197 g of Au contains=6.022×1023
Therefore 1g of Au contains=6.022×1023197×1=3.06×1021atoms
(ii) Gram atomic mass of Li=7g
23g of Na contains atom=6.022×1023
1g of Na contains atoms=6.022×10237×1=86.0×1021atoms
Hence 1g of Li has a large number of atoms
(2)Moles of ethanol=0.04mol
Moles of water=1−0.04=0.96mol
Mass of water=0.96×18=17.3g
Mass of ethanol=0.04×46=1.84g
Mass of solution=17.3+1.84=19.1g
Let the density of solution be 1g/ml
Volume of solution=massdensity=19.11=19.1ml=0.0191L
Molarity =0.04/0.0191=2.1M