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The density of a DNA sample is \(1.1 \mathrm{~g} / \mathrm{ml}\) and its molar mass determined by cryoscopic method was found to be \(6 \times 10^{8} \mathrm{~g} / \mathrm{mole}\). What is the volume occupied by one DNA molecule? \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(5.45 \times 10^{8} \mathrm{ml}\) (b) \(1.83 \times 10^{-9} \mathrm{ml}\) (c) \(9.06 \times 10^{-16} \mathrm{ml}\) (d) \(1.09 \times 10^{-13} \mathrm{ml}\)

Short Answer

Expert verified
\(1.83 \times 10^{-16} \mathrm{ml}\)

Step by step solution

01

Calculate the number of moles in one DNA molecule

Since one mole of DNA has a mass of \(6 \times 10^{8}\) grams, and Avogadro's number \(N_{A}\) is \(6 \times 10^{23}\), the mass of one DNA molecule can be found by dividing the molar mass by Avogadro's number.
02

Calculate the mass of one DNA molecule

Divide the molar mass by Avogadro's number to find the mass of a single DNA molecule: \(\text{mass of one DNA molecule} = \frac{6 \times 10^{8}\text{ g/mole}}{6 \times 10^{23}\text{ molecules/mole}} = 1 \times 10^{-15}\text{ g/molecule}\).
03

Calculate the volume occupied by one DNA molecule

Using the density formula \(\text{density} = \frac{\text{mass}}{\text{volume}}\), rearrange to solve for volume: \(\text{volume} = \frac{\text{mass}}{\text{density}}\). Then substitute the mass of one DNA molecule and density of the DNA sample: \(\text{volume} = \frac{1 \times 10^{-15}\text{ g}}{1.1\text{ g/ml}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
The molar mass is a fundamental concept in chemistry that denotes the mass of a substance (typically in grams) per mole of its entities (such atoms, molecules, or ions). To understand it with an example, consider table salt (NaCl). Its molar mass is the sum of the atomic mass of sodium (Na, approximately 23 g/mol) and chlorine (Cl, approximately 35.5 g/mol), which totals roughly 58.5 g/mol. This is the mass of one mole of NaCl.

When dealing with complex molecules like DNA, the molar mass can be significantly higher, as in the exercise where it is given as \(6 \times 10^{8} \mathrm{~g/mole}\). This huge number represents the combined mass of all the atoms in one mole of DNA molecules. Having the molar mass allows understanding of the amount of substance and further enables the calculation of the number of particles in a sample when combined with Avogadro's number.
Cryoscopic Method
The cryoscopic method is a technique used to determine the molar mass of a substance by measuring the depression in freezing point of a solvent when the substance is dissolved in it. This method is based on the colligative properties of solutions, which are properties that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, rather than the identity of the particles.

In cryoscopy, the freezing point of the pure solvent is compared to the freezing point of the solution. From the depression in freezing point, and by knowing the proportionality constant (cryoscopic constant) of the solvent, the molar mass of the solute can be calculated. This is particularly useful for large complex molecules such as proteins or DNA where other methods of molar mass determination might be challenging.
Avogadro's Number
Avogadors's Number, denoted as \(N_{A}\), is a fundamental physical constant that is the cornerstone of chemistry. It is the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Avogadro's number is approximately \(6.022 \times 10^{23}\) entities per mole.

This constant is crucial when converting between the number of atoms or molecules and the amount of substance in moles. For instance, if we have one mole of DNA molecules, that's equivalent to approximately \(6 \times 10^{23}\) DNA molecules, based on Avogadro's number. This also aids in the precise calculation of the mass of an individual DNA molecule by dividing the molar mass by Avogadro's number, as shown in the provided exercise solution.
Density Formula
We often need to calculate the amount of space a substance occupies, and that's where the density formula comes into play. Density is defined as the mass per unit volume of a material and is expressed as \(density = \frac{mass}{volume}\). It's a critical physical property that can be used to identify substances or to calculate either mass or volume if the other is known.

In the context of the DNA molecular volume calculation, we already have the mass of a single DNA molecule and the density of the DNA sample. To find the volume occupied by one DNA molecule, one needs to rearrange the density formula to solve for volume, which becomes \(volume = \frac{mass}{density}\). By inserting the known values into the rearranged formula, the volume occupied by a single DNA molecule can be accurately calculated. In the given exercise, using the density of DNA and the mass of one DNA molecule, derived from its molar mass and Avogadro's number, we find the volume a single DNA molecule occupies.

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Most popular questions from this chapter

A certain vitamin extracted from plant sources has carbon and hydrogen in \(8: 1\) mass ratio. The percentage of oxygen is nearly 7.3. The compound gave no test for nitrogen or sulphur or any other element. What should be the empirical formula of the compound? (a) \(\mathrm{C}_{30} \mathrm{H}_{45} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{15} \mathrm{H}_{23} \mathrm{O}\) (c) \(\mathrm{C}_{29} \mathrm{H}_{45} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{O}\)

Hydrogen cyanide, \(\mathrm{HCN}\), can be made by a two-step process. First, ammonia is reacted with \(\mathrm{O}_{2}\) to give nitric oxide, \(\mathrm{NO}\). \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Then nitric oxide is reacted with methane, \(\mathrm{CH}_{4}\) \(2 \mathrm{NO}(\mathrm{g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \rightarrow 2 \mathrm{HCN}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(+\mathrm{H}_{2}(\mathrm{~g})\) When \(25.5 \mathrm{~g}\) of ammonia and \(32.0 \mathrm{~g}\) of methane are used, how many grams of hydrogen cyanide can be produced? (a) \(1.5\) (b) \(2.0\) (c) \(40.5\) (d) \(54.0\)

The empirical formula of a compound is \(\mathrm{CH}_{2} \mathrm{O}\). If \(0.0833\) moles of the compound contains \(1.0 \mathrm{~g}\) of hydrogen, its molecular formula should be (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\)

Two successive reactions, \(\mathrm{A} \rightarrow \mathrm{B}\) and \(\mathrm{B} \rightarrow \mathrm{C}\), have yields of \(90 \%\) and \(80 \%\), respectively. What is the overall percentage yield for conversion of \(\mathrm{A}\) to \(\mathrm{C}\) ? (a) \(90 \%\) (b) \(80 \%\) (c) \(72 \%\) (d) \(85 \%\)

A hydrocarbon \(\mathrm{C}_{n} \mathrm{H}_{2 n}\) yields \(\mathrm{C}_{n} \mathrm{H}_{2 n+2}\) by reduction. In this process, the molar mass of the compound is raised by \(2.38 \%\). The value of \(n\) is (a) 8 (b) 4 (c) 6 (d) 5

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