The ratio of number of oxygen atoms (O) in 16.0 ~g ozone (O_3), 28.0 ~g carbon monoxide (CO) and 16.0 oxygen (O_2) is (Atomic mass : C=12, O=16 and Avogadro's constant N_ A =6.0 × 10^ 23 ~mol^ -1 )

Correct answer: D. 1: 1: 1 16.0 g O_3=(16)/(48) mole aligned & =(16)/(48) ×6.023 ×10^ 23 molecules & =3 × (16)/(48) ×6.023 ×10^ 23 atoms & =6.023 ×10^ 23 atoms & 28.0 g CO=(28)/(28) mole =1 mole & =1 ×6.023 ×10^ 23 molecules & =1 ×6.023 ×10^ 23 atoms & =6.023 ×10^ 23 atoms & 16.0 g O_ 2 =(16)/(16) mole =1 mole aligned =1 ×6.023 ×10^ 23 molecules =1 ×6.023 ×10^ 23 atoms =6.023 ×10^ 23 atoms Therefore, the ratio is 6.023 ×10^ 23 : 6.023 ×10^ 23 : 6.023 ×10^ 23 i.e. 1: 1: 1

JEE Main 20127 MayOnline

Exam: JEE Main
Year: 2012
Date: 7 May
Mode: Online
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2012 · 7 May · Online · Chemistry · Some Basic Concepts of Chemistry

The ratio of number of oxygen atoms (O) in 16.0 ~g ozone (O_3),...

The ratio of number of oxygen atoms

(\mathrm{O})

16.0 \mathrm{~g}

ozone \left(\mathrm{O}_3\right), 28.0 \mathrm{~g}

carbon monoxide

(\mathrm{CO})

and 16.0 oxygen \left(\mathrm{O}_2\right)

is (Atomic mass :

\mathrm{C}=12, \mathrm{O}=16

and Avogadro's constant

\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}

)

Options

A $3: 1: 2$
B $1: 1: 2$
C $3: 1: 1$
D $1: 1: 1$

Correct answer

D. $1: 1: 1$

Step-by-step solution

16.0 \mathrm{ g} \mathrm{O}_3=\frac{16}{48}

mole

\begin{aligned}& =\frac{16}{48} \times6.023 \times10^{23} \text{ molecules }\\ & =3 \times \frac{16}{48} \times6.023 \times10^{23} \text{ atoms }\\ & =6.023 \times10^{23} \text{ atoms }\\ & 28.0 \mathrm{ g} \mathrm{CO}=\frac{28}{28} \text{ mole }=1 \text{ mole }\\ & =1 \times6.023 \times10^{23} \text{ molecules }\\ & =1 \times6.023 \times10^{23} \text{ atoms }\\ & =6.023 \times10^{23} \text{ atoms }\\ & 16.0 \mathrm{ g} \mathrm{O}_{2}=\frac{16}{16} \text{ mole }=1 \mathrm{ mole} \end{aligned}

=1 \times6.023 \times10^{23}

molecules

=1 \times6.023 \times10^{23}

atoms

=6.023 \times10^{23}

atoms Therefore, the ratio is

6.023 \times10^{23}: 6.023 \times10^{23}: 6.023 \times10^{23}

i.e.

1: 1: 1$

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