Ncert Exercises & Solutions

1.28 Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl2(g)

(i). 1 g of Au (s) = 1/197 mol of Au (s)
= $\frac{6.022 x 10^{23}}{197}$ atoms of Au (s)
= 3.06 × $10^{21}$ atoms of Au (s)

(ii). 1 g of Na (s) = 1/23 mol of Na (s)
= $\frac{6.022 x 10^{23}}{23}$ atoms of Na (s)
= 0.262 × $10^{23}$ atoms of Na (s)
= 26.2 × $10^{21}$ atoms of Na (s)

(iii). 1 g of Li (s) = 1/7 mol of Li (s)
= $\frac{6.022 x 10^{23}}{7}$ atoms of Li (s)
= 0.86 × $10^{23}$ atoms of Li (s)
= 86.0 × $10^{21}$ atoms of Li (s)

(iv). 1 g of Cl2 (g) = 1/71 mol of Cl2 (g)

(Molar mass of Cl2 molecule = 35.5 × 2 = 71 g mol–1)
= $\frac{6.022 x 10^{23}}{71}$ atoms of Cl2 (g)
= 0.0848 × $10^{23}$ atoms of Cl2 (g)
= 8.48 × $10^{21}$ atoms of Cl2 (g)
Hence, 1 g of Li (s) will have the largest number of atoms.

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