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Kinetic Theory of Gases

The ratio of ...

Question

The ratio of molecules present in $4.4$ g of $C O_{2}$ and $0.56$ L (at STP) of $S O_{2}$ is:
(Atomic mass: $C = 12, 0 = 16, S = 32$ g/mole)

A

2 : 1

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B

4 : 1

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C

8 : 1

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D

3 : 2

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Solution

The correct option is B $4 : 1$
Moles of $C O_{2}$ = $\frac{4.4}{44} = 0.1$

Molecules of $C O_{2} = 0.1 N_{A}$

Moles of $S O_{2} = \frac{0.56}{22.4} = 0.025$

Molecules of $S O_{2} = 0.025 N_{A}$

So the ratio $= 4 : 1$
Option B is correct.

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