Question

The average density of the universe as a whole is estimated as \[3 \times {10^{29}}\] g/ml. If we assume that the entire mass is only H atoms, what is the average volume of space that contains once H atoms?
A.111.111 l
B.\[1.8 \times {10^{ - 4}}\]l
C.\[3.6 \times {10^{ - 4}}\]l
D.55.56 l

Answer 1

Hint:The term "density" refers to the amount of mass per unit of volume. An object's average density is proportional to its total mass divided by its total volume. An object composed of a more dense material (such as iron) would have less volume than one made of a less dense substance of comparable mass (such as water).

Complete answer:
Density of universe =\[3 \times {10^{29}}\] g/ml
We know that Atomic mass of Hydrogen = 1 amu
To find the mass of one Hydrogen atom, we divide it using Avogadro's number.
Mass of one Hydrogen atom = \[\dfrac{1}{{6.023 \times {{10}^{23}}}}\]
Mass of one Hydrogen atom = \[1.66 \times {10^{ - 24}}\]
​ The term "density" refers to the amount of mass per unit of volume. An object's average density is proportional to its total mass divided by its total volume. An object composed of a more dense material (such as iron) would have less volume than one made of a less dense substance of comparable mass (such as water).
\[d = \dfrac{m}{v}\], where m = Mass, v = Volume
\[v = \dfrac{m}{d}\]
Substituting the values, we get
\[v = \dfrac{{1.66 \times {{10}^{ - 24}}}}{{3 \times {{10}^{ - 29}}g/ml}}\]
\[v = 0.5533 \times {10^5}\]
​v = 55333.33 ml
v = 55.56 litre

Hence, Option D is correct.

Note:
The term "density" refers to the amount of mass per unit of volume. An object's average density is proportional to its total mass divided by its total volume. An object composed of a more dense material (such as iron) would have less volume than one made of a less dense substance of comparable mass (such as water).