The average density of the universe as a whole is estimated at $3 \times 10^{- 29} \frac{g}{m L}$ . If we assume that the entire mass under consideration is only that of H atoms, what is the average volume of space that contains one H atom?

111.111

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1.8 \times 10^{- 5}

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55.55

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3.6 \times 10^{- 5}

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Solution

The correct option is C $55.55$ L
Density is given by the ratio of mass to volume. Here, volume corresponding to one H-atom and density is provided.
Number of moles corresponding to one H-atom is given by $\frac{1}{N_{A}}$ mol.
$∴ M a s s o f H - a t o m (m) = m o l e s \times M o l a r M a s s$
$∴ M a s s o f H - a t o m (m) = \frac{1}{N_{A}} \times 1 = \frac{1}{N_{A}} g$
Also, $D e n s i t y (d) = \frac{m a s s (m)}{V o l u m e (V)}$
$∴ V o l u m e (V) = \frac{m a s s (m)}{D e n s i t y (d)}$
$∴ V o l u m e (V) = \frac{1}{N_{A} \times (3 \times 10^{- 29})}$
$∴ V o l u m e (V) = 55555.55 m L = 55.55 L$

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