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NCERT Class XI Chemistry Some Basic Concepts of Chemistry Solutions

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Question : 28 of 36
Marks: +1, -0
Which one of the following will have largest number of atoms?
(i) 1 g Au(s)\mathrm{Au}_{(s)}
(ii) 1 g Na(s)\mathrm{Na}_{(s)}
(iii) 1 g Li(s)\mathrm{Li}_{(s)}
(iv) 1 g of Cl2(g)\mathrm{Cl}_{2(g)}
Solution:  
(i) No. of atoms in 1 g of Au = 1197\frac{1}{197} × 6.022 × 102310^{23} = 3.057 × 102110^{21} atoms
(ii) No. of atoms in l g of Na = 123\frac{1}{23} × 6.022 × 102310^{23} = 2.618 × 102210^{22} atoms
(iii) No. of atoms in l g of Li = 17\frac{1}{7} × 6.022 × 102310^{23} = 8.604 × 102210^{22} atoms
(iv) No. of atoms in 1 g of Cl2\mathrm{Cl}_2 = 171\frac{1}{71} × 2 × 6.022 × 102310^{23} = 1.697 × 102210^{22} atoms
Thus, 1 g of Li has largest number of atoms.
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