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The ratio of molecules present in $16 g$ of methane ( ${CH}_{4}$ ) and $16 g$ of oxygen ( $O_{2}$ ) is

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2:1

1:2

2:3

3:2

answer is A.

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Detailed Solution

The ratio of molecules present in

16 g

of methane and

16 g

of oxygen is 2:1.
Number of molecules

=

\frac{Given mass}{Molecular mass} \times Avogadro number

Number of methane molecules in,

16 g =

\frac{16}{16} \times N_{A}

\Rightarrow = N_{A}

Number of oxygen molecules in,

16 g =

\frac{16}{32} \times N_{A}

\Rightarrow = \frac{N_{A}}{2}

Ratio of molecules present in

16 g

of methane (

{CH}_{4}

) and

16 g

of oxygen (

O_{2}

)

= \frac{N_{A} \times 2}{N_{A}}

\Rightarrow \frac{2}{1}

\Rightarrow 2 : 1

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