Question

The average density of the universe as a whole is estimated as \(3 \times 10^{-29} \mathrm{~g} / \mathrm{ml}\). If we assume that the entire mass is only \(\mathrm{H}\) atoms, what is the average volume of space that contains one \(\mathrm{H}\) atom? (a) \(111.111\) (b) \(1.8 \times 10^{-5} 1\) (c) \(55.561\) (d) \(3.6 \times 10^{-5} 1\)

Short answer

The correct average volume that contains one hydrogen atom is approximately 5.535 \(\times\) 10^24 liters, which is not present in the given options; thus, the question may contain an error or require unit conversion.

Step-by-step solution

Understand the problem

We are given the average density of the universe and are asked to find the average volume of space for one hydrogen atom (H atom). Since the density is given in grams per milliliter (g/ml) and we are treating the mass of the universe as hydrogen atoms, we can use the mass of a single hydrogen atom to find the volume that one atom occupies.

Find the mass of a hydrogen atom

The mass of one hydrogen atom can be found using the molar mass of hydrogen and Avogadro's number. The molar mass of hydrogen is approximately 1 gram per mole, and Avogadro's number is approximately 6.022 \(\times\) 10^23 atoms per mole. The mass of one hydrogen atom is therefore 1 g divided by 6.022 \(\times\) 10^23 atoms/mole.

Calculate the volume for one hydrogen atom

Using the density formula D = m/V where D is density, m is mass, and V is volume, we can solve for V to find the volume of one hydrogen atom. We rearrange the formula to V = m/D. Substituting the values, we have V = (1 g / 6.022 \(\times\) 10^23 atoms/mole) / (3 \(\times\) 10^-29 g/ml), which will give us the volume in milliliters.

Perform the calculation

We calculate the volume V as follows: V = (1 / 6.022 \(\times\) 10^23) / (3 \(\times\) 10^-29) = (1 / 6.022) \(\times\) (10^23 / 10^-29) \(\times\) (1 / 3) ml = (1.6605 \(\times\) 10^-24) \(\times\) 10^52 \(\times\) (1 / 3) ml = (1.6605 / 3) \(\times\) 10^28 ml

Simplify the result

Upon simplification, we have V \(\approx\) (0.5535) \(\times\) 10^28 ml or V \(\approx\) 5.535 \(\times\) 10^27 ml. Convert milliliters to liters by dividing by 1000, since there are 1000 milliliters in a liter. V \(\approx\) 5.535 \(\times\) 10^24 liters. This result must match one of the given option units for consistency.

Key concepts

Density of the Universe

When we talk about the density of the universe, we're trying to understand how much mass exists per unit volume in the vast expanse of space. It's a fundamental concept in cosmology and helps us derive many aspects of the universe's structure and evolution.

The average density is incredibly low because space is mostly empty, with vast distances between stars and galaxies. In scientific terms, it's often expressed in grams per cubic centimeter or milliliters. The estimated value of the universe's average density is about \(3 \times 10^{-29}\) grams per milliliter.

Understanding this density is crucial since it influences the universe's expansion rate and fate. If the density is above a certain critical value, the gravitational pull could eventually halt expansion and lead to a contraction, or 'big crunch'. If it's below this critical value, the universe could continue expanding forever.

Molar Mass of Hydrogen

The molar mass of a substance is the mass of one mole of that substance. For hydrogen, the lightest and most abundant element in the universe, its molar mass is particularly simple to remember — it's approximately 1 gram per mole.

The molar mass is used to convert moles, a measurement of the amount of substance, to grams, a measurement of mass. Since hydrogen has one proton and often one electron, with negligible electron mass, its molar mass is essentially equal to the mass of its atomic nucleus, the proton.

This simple relationship allows for straightforward calculations when dealing with hydrogen in chemical equations and, as seen in the exercise, cosmological contexts. The molar mass serves as a bridge between the microscopic world of atoms and the macroscopic world of grams and kilograms that we interact with.

Avogadro's Number

Avogadro's number is a constant that represents the exact number of entities, like atoms or molecules, in one mole of a substance. It's named after the Italian scientist Amedeo Avogadro and is approximately \(6.022 \times 10^{23}\). This staggering number is essential in chemistry and physics because it allows us to link the atomic scale with the macroscopic scale.

For example, if we know the molar mass of an element or compound and how many moles we have, we can calculate the number of atoms or molecules present, and vice versa. Avogadro's number thus provides a way to relate the mass of a substance to the number of particles it contains, making it a fundamental cornerstone in calculations that involve the atomic or molecular scale.

Knowing Avogadro's number allows us to perform the calculations seen in the given exercise, as it's the link between the mass of a single atom, the molar mass, and the number of atoms in a given sample.