. Specific volume of cylindrical virus particle
is 6.02×10-2 cc/g whose radius and length are 7Å and 10 A respectively. If NA = 6.02×1023, find molecular weight of virus.
(a) 15.4 kg/mol
(b) 1.54×104 kg/mol
(c) 3.08×104 kg/mol
(d) 3.08×103 kg/mol
Best Answer
Correct Option is:
Specific volume (vol. of 1 g ) of cylindrical virus particle =6.02 × 10-2 cc/g
Radius of virus, r =7Å=7×10-8 cm
Length of virus, l = 10 10-8 cm
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Therefore, molecular weight of virus is 15.4 kg/mol
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