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CGPET 2011 Solved Paper

Section: Chemistry

Question : 82 of 150

Marks: +1, -0

Solution:

No. of atoms in

1\ \mathrm{g}

\mathrm{O}_2(\mathrm{g})\;=2 \times \;\frac{1}{32} \times 6.023 \times 10^{23}

\;=0.38 \times 10^{23}

No. of atoms in

1\ \mathrm{g}

\mathrm{Ni}(\mathrm{s})\;=\;\frac{1}{58.2} \times 6.023 \times 10^{23}

\;=0.10 \times 10^{23}

No. atoms in

1\ \mathrm{g}

\mathrm{B}(\mathrm{s})\;=\;\frac{1}{10.8} \times 6.023 \times 10^{23}

\;=0.56 \times 10^{23}

No. of atoms in

1\ \mathrm{g}

\mathrm{N}_2(\mathrm{g})\;=2 \times \;\frac{1}{28} \times 6.023 \times 10^{23}

\;=0.43 \times 10^{23}

Thus,

1\ \mathrm{g}\ \mathrm{B}(\mathrm{s})

contains the largest number of atoms.

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