Problem 35 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A plant virus is found to consist of uniform cylindrical particles of \(150 \AA\) in diameter and \(5000 \AA\) long. The specific volume of the virus is \(0.75 \mathrm{~cm}^{3} / \mathrm{g}\). If the virus is considered to be a single particle, find its molar mass.

Step-by-Step Solution

Answer

The molar mass of the virus is approximately \(7.10 \times 10^7 \text{ g/mol}\).

Step 1: Determine the Volume of a Single Virus Particle

The volume of a cylinder can be calculated using the formula \( V = \pi r^2 h \) where \( r \) is the radius and \( h \) is the height. Here, the diameter is given as \( 150 \AA \) so the radius \( r = \frac{150}{2} = 75 \AA \). Now calculate the volume \( V \) in \( \AA^3 \): \[ V = \pi (75)^2 (5000) \approx 8.85 \times 10^7 \AA^3 \].

Step 2: Convert Volume from Cubic Angstroms to Cubic Centimeters

To find the volume in cubic centimeters, use the conversion \( 1 \AA^3 = 10^{-24} \text{ cm}^3 \). Thus: \[ V = 8.85 \times 10^7 \AA^3 \times 10^{-24} \text{ cm}^3/\AA^3 = 8.85 \times 10^{-17} \text{ cm}^3 \].

Step 3: Calculate the Mass of a Single Virus Particle

From the specific volume of the virus given as \(0.75 \text{ cm}^3/\text{g} \), the mass \( m \) of a single virus particle is given by \( m = \frac{V}{ ext{specific volume}} \). Substitute the volume: \[ m = \frac{8.85 \times 10^{-17} \text{ cm}^3}{0.75 \text{ cm}^3/ ext{g}} = 1.18 \times 10^{-16} \text{ g} \].

Step 4: Calculate the Molar Mass of the Virus

The molar mass \( M \) can be found using Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \). The formula is \( M = m \times N_A \). Substitute the mass: \[ M = 1.18 \times 10^{-16} \text{ g} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 7.10 \times 10^7 \text{ g/mol} \].

Key Concepts

Cylindrical Particles

Cylindrical particles are common shapes found in many biological and physical structures. They resemble tiny cylinders and, in this context, we are examining virus particles viewed as uniform cylinders. Calculating the volume of such a cylindrical particle can be done easily using the formula:
\[ V = \pi r^2 h \]
Where:

• \( V \) is the volume of the cylinder.
• \( r \) is the radius of the base of the cylinder.
• \( h \) is the height or length of the cylinder.

In our example, the cylindrical virus particles have a diameter of \( 150\, \AA \) and a length of \( 5000\, \AA \). To calculate the volume, first find the radius by dividing the diameter by two. This gives us a radius of \( 75\, \AA \). With these values inserted into the formula, we determine the overall particle volume to be approximately \( 8.85 \times 10^7 \AA^3 \). Understanding how to compute volumes is critical in finding the masses and molar masses of biological entities shaped like cylinders.

Specific Volume

Specific volume is a key concept in determining the density of a substance. It is defined as the total volume per unit mass:
\[ \text{Specific Volume} = \frac{V}{m} \]
Where:

• \( V \) is the volume.
• \( m \) is the mass.

In the given problem, the virus's specific volume is \( 0.75\, \text{cm}^3/\text{g} \). This means for each gram of virus, the total volume occupied is \( 0.75\, \text{cm}^3 \). To ascertain the mass of one virus particle, divide the particle's volume by the specific volume. After converting the volume from cubic angstroms to cubic centimeters (\( 8.85 \times 10^{-17} \text{ cm}^3 \)), use the specific volume to get the mass:
\[ m = \frac{8.85 \times 10^{-17} \text{ cm}^3}{0.75 \text{ cm}^3/\text{g}} \approx 1.18 \times 10^{-16} \text{ g} \].
This mass value is essential for calculating the molar mass, which indicates how many grams one mole of these viral particles would contain.

Avogadro's Number

Avogadro's number is critical in chemistry for converting between atoms or molecules and moles. It is defined as the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Avogadro's number \( N_A \) is approximately \( 6.022 \times 10^{23} \text{ mol}^{-1} \).
In our exercise, we use Avogadro's number to find the molar mass of virus particles. The formula to find the molar mass \( M \) given the mass of a single particle \( m \) is:
\[ M = m \times N_A \]
Substitute the mass from the problem, \( 1.18 \times 10^{-16} \text{ g} \), to find:
\[ M = 1.18 \times 10^{-16} \text{ g} \times 6.022 \times 10^{23} \text{ mol}^{-1} \approx 7.10 \times 10^7 \text{ g/mol} \].
This value of molar mass tells us how much one mole of these virus particles would weigh, providing a bridge between the microscopic scale and quantities we can observe and measure.