Question

The specific volume of cylindrical virus particle is $6.023 \times {10^{ - 2}}$cc/gm whose radius and length are $7\text{ }A{}^\circ \text{ }\!\!~\!\!\text{ }$ and $10\text{ }A{}^\circ \text{ }\!\!~\!\!\text{ }$ respectively. If ${N_A} = 6.02 \times {10^{23}}$, find the molecular weight of the virus.

A. $1.54Kg/mol$
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: Specific volume of any particle is defined as the number of cubic meters occupied by one gram of a particular. Mathematically it is defined as the ratio of a substance’s volume to its mass. The molecular weight of the virus is found by calculating the molecular weight of ${N_A}$ particles.

Complete step by step answer:
The specific volume of the cylindrical virus particle is $6.023 \times {10^{ - 2}}$cc/gm and the radius of the virus is $7\text{ }A{}^\circ \text{ }\!\!~\!\!\text{ }$ and the length of the virus particle is$10\text{ }A{}^\circ \text{ }\!\!~\!\!\text{ }$


$
  10 \times {10^{ - 8}}cm \\
   \Rightarrow {10^{ - 7}}cm \\
$
As we know the formula used for finding the volume of the cylindrical particle is $\pi {r^2}l$
Where π = $\dfrac{{22}}{7}$ and r is the radius and l is the length of the cylinder.
On putting the value of r and l in the above formula we get:
The volume of the cylindrical virus =$
  \dfrac{{22 \times {{(7 \times {{10}^{ - 8}})}^2} \times {{10}^{ - 7}}}}{7} \\
    \\
$
On solving we get the volume of cylindrical virus = $1.54 \times {10^{ - 21}}c{m^3}$
Since one mole of particles contains $6.022 \times {10^{23}}$ particles.
So the volume of one mole of cylindrical virus particle = ${N_A} \times 1.54 \times {10^{ - 21}}c{m^3}$ $
   \Rightarrow 6.022 \times {10^{23}} \times 1.54 \times {10^{ - 21}} \\
   \Rightarrow 9.27388 \times {10^2}c{m^3} \\
$
Molecular weight=$\dfrac{{volume{\text{ }}of{\text{ }}1mole}}{{specific{\text{ }}volume}}$
On putting values we get:
Molecular weight = $\dfrac{{9.27388 \times {{10}^2}}}{{6.022 \times {{10}^{ - 2}}}} = 1.54 \times {10^4}g/mol$
As we know 1 kg = 1000 g
Then Molecular Weight is given as:
$
  \dfrac{{1.54 \times {{10}^4}}}{{1000}} \\
   \Rightarrow \dfrac{{154 \times {{10}^4}}}{{1000 \times 10}} = 154kg/mol \\
$


So, the correct answer is Option A .

Note:
A molecular mass of any compound is defined as the mass of the molecule which is equal to the sum of masses of all the atoms contained in molecules. Frequently molecular weights are dimensionless, but they can be represented in kg/mol depending upon the method used for measurement.