The number of oxygen atoms in 6.4 g of oxygen are :

6.023 \times 10^{23}

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6.023 \times 10^{22}

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1.2046 \times 10^{23}

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1.2046 \times 10^{22}

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Solution

The correct option is B $1.2046 \times 10^{23}$
Number of moles of $O_{2} = \frac{6.4}{32} = 0.1 m o l e$
Number of $O_{2}$ atoms $= 0.2 \times 6.023 \times 10^{23}$ $= 1.2046 \times 10^{23}$

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Question 9
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68 \times 10^{23}$
(b) $6.09 \times 10^{22}$
(c) $6.022 \times 10^{23}$
(d) $6.022 \times 10^{21}$

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The number of oxygen atoms in 6.4 g of oxygen are :

The correct option is B 1.2046×1023Number of moles of O2=6.432=0.1 moleNumber of O2 atoms =0.2×6.023×1023=1.2046×1023

The correct option is B $1.2046 \times 10^{23}$
Number of moles of $O_{2} = \frac{6.4}{32} = 0.1 m o l e$
Number of $O_{2}$ atoms $= 0.2 \times 6.023 \times 10^{23}$ $= 1.2046 \times 10^{23}$