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Question

1) Which one having larger no of atom
i) 1 g Au
ii) 1g Li
2) Calculate the molarity of solution of ethanol in water in which mole fraction of ethanol is 0.040.

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Solution

(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=10.04=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

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