Question

Which has the minimum number of atoms among the following?
32 g of O2
33.6 L of CO2 at STP
3 moles of H2
O 2 mole of SO2​

Answer 1

Answer:

3 moles of H2 is the correct answer

Answer 2

Explanation:

(i) 32 g of $O_{2}$

1 mole of $O_{2} = 32 g$

1 mole of $O_{2} = 6.022*10^{23}$ molecules

1 molecule of $O_{2} = 2$ atoms of $O_{2}$

32 g of $O_{2} = 2 * 6.022*10^{23}$ atoms

32 g of  $O_{2} = 12.044*10^{23}$ atoms

(ii) 22.4 L of $CO_{2}$ at STP = 1 mole of $CO_{2}$

33.6 L of $CO_{2}$ at STP = $\frac{33.6}{22.4}$ mole

33.6 L of $CO_{2} = 1.5$ mole

1 mole of $CO_{2} =$$6.022*10^{23}$ molecules

1.5 moles of $CO_{2} =$ $1.5*6.022*10^{23}$ molecules

1.5 moles of $CO_{2} =$$9.033*10^{23}$ molecules

1 molecule of $CO_{2} = 3$ atoms

$9.033*10^{23}$ molecules of $CO_{2} = 3*9.033*10^{23}$ atoms

22.4 L of $CO_{2} = 27.099*10^{23}$ atoms

(iii) 1 mole of $H_{2}$ $= 6.022*10^{23}$ molecules

3 moles of $H_{2} = 3*6.022*10^{23}$ molecules

3 moles of $H_{2} = 18.066*10^{23}$ molecules

1 molecule of $H_{2} = 2$ atoms

$18.066*10^{23}$ molecules of $H_{2} = 2*18.066*10^{23}$ atoms

3 moles of $H_{2} = 36.132*10^{23}$ atoms

(iv) 1 mole of $SO_{2} = 6.022*10^{23}$ molecules

2 moles of $SO_{2} = 2* 6.022*10^{23}$ molecules

2 moles of $SO_{2} = 12.044*10^{23}$ molecules

1 molecule of $SO_{2} = 3$ atoms

$12.044*10^{23}$ molecules of $SO_{2} = 3*12.044*10^{23}$ atoms

2 moles of $SO_{2} = 36.132*10^{23}$ atoms

Hence, 32 g of $O_{2}$ has the minimum number of atoms among the given molecules.