Question

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

Answer 1

(i)Gram atomic mass of $Au=197g$

197 g of Au contains=$6.022\times {10}^{23}$

Therefore 1g of Au contains=$\frac{6.022\times {10}^{23}}{197\times 1}=3.06\times {10}^{21}atoms$

(ii) Gram atomic mass of $Li$=$7g$

23g of Na contains atom=$6.022\times {10}^{23}$

1g of Na contains atoms=$\frac{{6.022\times 10}^{23}}{7\times 1}=86.0{\times 10}^{21}atoms$

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=$0.96\times 18=17.3g$

Mass of ethanol=$0.04\times 46=1.84g$

Mass of solution=$17.3+1.84=19.1g$

Let the density of solution be 1g/ml

Volume of solution=$\frac{mass}{density}=\frac{19.1}{1}=19.1ml=0.0191L$

Molarity =$0.04/0.0191=2.1M$