Specific volume of cylinfrical virus particle is 6.02 $$ \times $$ 10$$-$$2 cc/g whose radius and length are 7 $$\mathop A\limits^ \circ $$ and 10 $$\mathop A\limits^ \circ $$ respectively. If NA = 6.02 $$ \times $$ 1023, find molecular weight of virus.
A
15.4 kg/mol
B
1.54 $$ \times $$ 104 kg/mol
C
3.08 $$ \times $$ 104 kg/mol
D
3.08 $$ \times $$ 103 kg/mol
Explanation
Specific volume ( vol. of 1 g) cylindrical virus particle