Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

Specific volume of cylindrical virus particle is \(6.02 \times\) \(10^{-2} \mathrm{cc} / \mathrm{g}\). Whose radius and length are \(7 \AA\) and \(10 \AA\) respectively. If \(\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23}\), find molecular weight of virus (a) \(15.4 \mathrm{~kg} / \mathrm{mol}\) (b) \(1.54 \times 10^{4} \mathrm{~kg} / \mathrm{mol}\) (c) \(4.68 \times 10^{4} \mathrm{~kg} / \mathrm{mol}\) (d) \(2.08 \times 10^{3} \mathrm{~kg} / \mathrm{mol}\)

Step-by-Step Solution

Answer

The molecular weight of the virus is (a) 15.4 kg/mol.

Step 1: Convert Angstroms to Centimeters

The dimensions of the virus are given in Angstroms. Convert them into centimeters. We know that 1 Angstrom (\(\AA\)) equals \(10^{-8}\) centimeters (cm). Thus, the radius \(r = 7 \AA\) becomes \(7 \times 10^{-8} \text{ cm}\) and the length \(h = 10 \AA\) becomes \(10 \times 10^{-8} \text{ cm}\).

Step 2: Calculate the Volume of the Cylinder

The volume \(V\) of a cylindrical virus can be calculated using the formula \(V = \pi r^2 h\). Substituting the radius \(r = 7 \times 10^{-8}\) cm and the height \(h = 10 \times 10^{-8}\) cm, we get:\[ V = \pi (7 \times 10^{-8})^2 (10 \times 10^{-8}) \]Calculating this yields:\[ V = \pi \times 49 \times 10^{-16} \times 10 = 490\pi \times 10^{-24} \text{ cubic centimeters} \]

Step 3: Determine the Mass of the Virus

Using the specific volume \(SV = \frac{V}{m}\) provided as \(6.02 \times 10^{-2} \text{ cc/g}\), solve for the mass \(m\) of one virus particle. We know:\[ \frac{V}{m} = 6.02 \times 10^{-2} \]Substituting the volume calculated, we get:\[ m = \frac{490 \pi \times 10^{-24}}{6.02 \times 10^{-2}} \]This simplifies to:\[ m = \frac{490 \pi \times 10^{-24}}{6.02 \times 10^{-2}} \approx 2552 \times 10^{-24} \text{ grams} \]

Step 4: Calculate Molecular Weight

The molecular weight \(M\) is the mass of one mole of virus particles divided by Avogadro's number \(N_A = 6.02 \times 10^{23}\). Therefore:\[ M = m \times N_A = 2552 \times 10^{-24} \times 6.02 \times 10^{23} \]This simplifies to:\[ M = 15.4 \text{ kg/mol} \]

Step 5: Choose the Correct Option

From the options provided, (a) \(15.4 \text{ kg/mol}\) matches our calculated molecular weight of the virus.

Key Concepts

Cylinder Volume Calculation

The calculation of the volume of a cylinder is an essential skill in physics and chemistry, especially when dealing with objects that have a cylindrical shape, like certain viral particles. To find the volume of a cylinder, you need to know the formula:\[ V = \pi r^2 h \]where:

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

In the case of the virus analyzed in the exercise, its dimensions were given in units of angstroms, which needed to be first converted into centimeters. Using the converted measurements, the radius \( r = 7 \times 10^{-8} \text{ cm} \) and the height \( h = 10 \times 10^{-8} \text{ cm} \), we applied the formula to determine the cylinder's volume. The multiplication of these values shows how the dimensions of an object can deeply influence the resulting volume, and understanding this process is key for accurate calculations in scientific studies.

Specific Volume

Specific volume is a concept that relates the volume of a substance to its mass, and it is expressed in cubic centimeters per gram (\(\text{cc/g}\)). In simple terms, it's a way to understand how much space a certain mass of substance occupies. The formula is given by:\[ SV = \frac{V}{m} \]where:

• \( SV \) is the specific volume
• \( V \) is the volume
• \( m \) is the mass

In the problem, the specific volume for the virus was given as \(6.02 \times 10^{-2} \text{ cc/g}\). Using this value, along with the calculated volume from the previous section, one can determine the mass of an individual virus particle. By reversing the formula, our goal was to solve for the mass \( m \) using the known specific volume and the calculated volume, which is crucial in finding other physical properties such as molecular weight.

Angstrom to Centimeter Conversion

Conversion of units is a fundamental aspect in scientific calculations to ensure consistency and accuracy. One common conversion involves angstroms (\( \AA \)), often used in chemistry and physics to express atomic-scale distances. To convert angstroms to centimeters, remember that:

• 1 angstrom = \( 10^{-8} \) centimeters

For example, in the exercise, the radius and height of the virus were given as 7 angstroms and 10 angstroms respectively. These were converted to centimeters before performing the cylinder volume calculations:- Radius: \( 7 \AA = 7 \times 10^{-8} \text{ cm} \)- Height: \( 10 \AA = 10 \times 10^{-8} \text{ cm} \)Making these conversions correctly is pivotal, as using incorrect units can result in significant calculation errors, especially when dealing with small-scale measurements such as viral particle dimensions.

Viral Particle Mass Determination

Determining the mass of a viral particle involves using its specific volume and calculated volume. Initially, we computed the volume of the virus from its cylindrical shape. Then, knowing its specific volume helps to deduce the mass using the equation:\[ m = \frac{V}{SV} \]In the exercise, after substituting the calculated volume and given specific volume, we find that the mass of a single virus particle is approximately \( 2552 \times 10^{-24} \text{ grams} \).Finally, to determine the molecular weight, which is the mass of one mole of these particles, it's necessary to multiply the virus's mass by Avogadro's number \( N_A = 6.02 \times 10^{23} \). This leads us to the molecular weight calculation, yielding \( M = 15.4 \text{ kg/mol} \). Understanding this process is vital for calculations related to molar quantities in chemistry.