Question

Specific volume of cylindrical virus particles is $6.02\times {{10}^{-2}}cc/gm$ whose radius and length are $7\overset{\circ }{\mathop{A}}\,$ and $10\overset{\circ }{\mathop{A}}\,$, respectively. If ${{N}_{A}}=6.02\times {{10}^{23}}$ find the molecular weight of the virus.
a.) $15.4kg/mol$
b.) $1.54\times {{10}^{4}}kg/mol$
c.) $3.08\times {{10}^{4}}kg/mol$
d.) $1.54\times {{10}^{3}}kg/mol$

Answer 1

Hint: We know that specific volume of a substance is the ratio of the substance’s volume to mass of the substance and we can use this definition of specific volume for finding a solution to this problem.

Complete step-by-step answer:
Given, that specific volume of one virus particle = $6.02\times {{10}^{-2}}cc/gm$
Here, volume of virus particle = volume of cylinder
And we that volume of a cylinder $=\pi {{r}^{2}}h$
Then by putting values of all the given parameters,
Volume of virus particle $=3.14\times {{7}^{2}}\times 10$
$=1538.6\times {{10}^{-30}}{{m}^{3}}$
$=1.54\times {{10}^{-27}}{{m}^{3}}$
Then we have to calculate volume of 1 mole virus particle $=6.023\times {{10}^{23}}\times 1.54\times {{10}^{-27}}{{m}^{3}}$
$=9.27\times {{10}^{-4}}{{m}^{3}}$
$=9.27\times {{10}^{2}}cc$
And, we know that specific volume = volume of 1 mol virus/ molecular weight
Then the molecular weight =volume of 1 mole virus / specific volume
$=\dfrac{9.27\times {{10}^{2}}}{6.02\times {{10}^{-2}}}$
$=1.54\times {{10}^{4}}g$
Here, we know that this mass is for 1 mole virus particles, and also we know that molecular weight is also of 1 mole substance.
Weight of 1 mole of virus particles $=1.54\times {{10}^{4}}g$
So, the molecular weight $=1.54\times {{10}^{4}}g/mol$ =$15.4kg/mol$

So, the correct answer is “A”.

Note: In this problem we should be careful about the units and conversion of units into one another, because there are a lot of unit conversions. Here the volume of cylindrical virus (for one virus) is given, not the volume of one cylinder which contains a lot of viruses.