gneet

Question
Q137) Specific volume of cylindrical virus particle is 6.02 × 10^{-2} cc/gm whose radius and length are 7 Å and 10 Å respectively.
If N_{A}= 6.02 ×10^{23}, find molecular weight of virus.

SOLUTION

W is the weight of one virus

∴molecular weight M of virus = w ×NA

Answer : (a)

1 Å = 10^{-10} m

Volume of virus = πr^{2}h

= π ×(7×10^{-10})2× 10×^{-10} cc

= 1.54×10^{-27} m^{3}

6.02× 10^{-2} cc = 1 g

Or 6.02 × 10^{-8} m^{3} = 10^{-3} kg

∴ weight of 1.54 × 10^{-27} m^{-} = w

W is the weight of one virus

∴molecular weight M of virus = w ×NA

Answer : (a)