Question

Specific volume of cylindrical virus particle is $6.02 \times {10^{ - 2}}\,cc/gm$ whose radius and length are $7\,\mathop {\text{A}}\limits^ \circ $ and $10\,\mathop {\text{A}}\limits^ \circ $ respectively. If ${N_A} = 6.02 \times {10^{23}}$. Find the molecular weight of the virus.
A.$15.4\,kg/mol$
B.$1.54 \times {10^4}\,kg/mol$
C.$3.08 \times {10^4}\,kg/mol$
D.$3.08 \times {10^3}\,kg/mol$

Answer 1

Hint: We have to convert the angstrom into centimeters. From the given length and radius, we have to calculate the volume of the virus. From the calculated volume of the virus, we have to calculate the volume of one mole of virus using Avogadro’s number. From the volume of one mole of virus, we have to calculate the molecular weight using the volume of one mole and specific volume.
Formula used:
We can calculate the volume of virus using the formula,
$V = \pi {r^2}h$
Here, r is the radius of the virus and h is the length.
We can calculate the volume of one mole of virus using the formula,
Volume of one mole of virus=${N_A} \times Volume\,of\,1\,virus$
Here ${N_A}$ is Avogadro’s number
We can calculate the molecular weight of the virus using the formula,
Molecular weight=$\dfrac{{Volume\,of\,1\,mole\,of\,virus}}{{Specific\,volume}}$

Complete step by step answer:
Given,
The length of the virus (h) is ${\text{10}}\,\mathop {\text{A}}\limits^ \circ $.
The volume of the virus (V) is ${\text{7}}\,\mathop {\text{A}}\limits^{\text{o}} $.
The specific volume of the virus is $6.02 \times {10^{ - 2}}\,gm/cc$.
From the question, we can understand that the shape of the object is cylindrical and we know the volume of the cylinder. The expression to calculate the volume of the cylinder is,
$V = \pi {r^2}h$
Here, r is the radius of the virus and h is the length.
One angstrom is ${10^{ - 8}}\,cm$
As the virus is cylindrical, we can calculate the volume of virus using the formula,
$
  V = \pi {r^2}h \\
  V = \dfrac{{22}}{7}{\left( {7 \times {{10}^{ - 8}}cm} \right)^2}\left( {10 \times {{10}^{ - 8}}\,cm} \right) \\
  V = 1.54 \times {10^{ - 21}}\,c{m^3} \\
 $
The volume of the virus is$1.54 \times {10^{ - 21}}\,c{m^3}$.
We can now calculate the volume of one mole of virus by multiplying the volume of virus with Avogadro’s formula.
Volume of one mole of virus=${N_A} \times Volume\,of\,virus$
Volume of one mole of virus=$6.02 \times {10^{23}}\, \times 1.54 \times {10^{ - 21}}$
Volume of one mole of virus=$9.2708 \times {10^2}$
The volume of one mole of virus is $9.2708 \times {10^2}\,$.
From the calculated volume of one mole of virus, we can calculate the molecular weight of the virus as,
Molecular weight=$\dfrac{{Volume\,of\,1\,mole\,of\,virus}}{{Specific\,volume}}$
Molecular weight=\[\dfrac{{9.2708 \times {{10}^2}}}{{6.02 \times {{10}^{ - 2}}\,}}\]
Molecular weight=$1.54 \times {10^4}\,g/mol$
Converting the grams to kilograms, we get the molecular weight of the virus as,
Molecular weight (in kg)=$1.54 \times {10^4}\,\dfrac{g}{{mol}} \times \dfrac{{1\,kg}}{{1000\,g}}$
Molecular weight (in kg)=$15.4\,kg/mol$
The molecular weight of the virus is $15.4\,kg/mol$.
$\therefore $Option (A) is correct.

Note:
Molecular mass is known as molecular weight, which is the mass of a substance. The major difference between molecular mass and molar mass is that molar mass is mass of one mole of a specific substance but molecular mass is the mass of a molecule. Molecular mass could also be calculated by summing the atomic weight of the atoms present in the compound.
We can also calculate the molecular weight of the virus using the formula,
$M = \dfrac{V}{{1/l}}$