Some Basic Concepts Of Chemistry Question 70
Question: Specific volume of cylindrical virus particle is $ 6.02\times {10^{-2}}cc/g. $ whose radius and length are 7 $ \overset{o}{\mathop{A}}, $ & 10 $ \overset{o}{\mathop{A}}, $ respectively.
If $ N _{A}=6.02\times 10^{23}mo{l^{-1}} $ find molecular weight of virus
Options:
A) $ 3.08\times 10^{3}kg/mol $
B) $ 3.08\times 10^{4}kg/mol $
C) $ 1.54\times 10^{4}kg/mol $
D) $ 15.4kg/mol $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Specific volume (volume of 1 g) of cylindrical virus particle $ =6.02\times {10^{-2}}cc/g $ Radius of virus $ =7\overset{o}{\mathop{A}},=7\times {10^{-8}}cm $ Length of virus $ =10\times {10^{-8}}cm $ Volume of virus = $ \pi r^{2}I=\frac{22}{7}\times {{(7\times {10^{-~8}})}^{2}}\times 10\times {10^{-}}^{8} $ $ =154\times {10^{-23}}cc $
$ Wt.ofonevirusparticle=\frac{volume}{specificvolume} $
$ \therefore $ Mol. wt. of virus = Wt. of $ N _{A} $ particle $ =\frac{154\times {10^{-23}}}{6.02\times {10^{-2}}}\times 6.02\times 10^{23}=15400g/mol $
$ =15.4kg/mol $
