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.