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Question
16 g of oxygen has same number of molecules as in:
(i) \[\ce{16 g of CO}\]
(ii) \[\ce{28 g of N2}\]
(iii) \[\ce{14 g of N2}\]
(iv) \[\ce{1.0 g of H2}\]
Solution
(iii) \[\ce{14 g of N2}\]
(iv) \[\ce{1.0 g of H2}\]
Explanation:
(iii) The number of moles of \[\ce{N2}\] is calculated by using equation (1) as follows,
Moles of N2 = `(14 g)/((28 g)/(mol)` = 0.5 mol
The number of molecules can be calculated by using equation (2) as follows,
0.5 mol = `"Number of molecules"/(6.022 xx 10^23)`
Number of molecules = `0.5 xx 6.022 xx 10^23`
(iv) The number of moles of \[\ce{H2}\] is calculated by using equation (1) as follows,
Moles of H2 = `(1 g)/((2 g)/(mol)` = 0.5 mol
The number of molecules can be calculated by using equation (2) as follows,
0.5 mol = `"Number of molecules"/(6.022 xx 10^23)`
Number of molecules = `0.5 xx 6.022 xx 10^23`
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