1
Specific volume of cylindrical virus particle is 6.02 10-2 cc/g whose radius and length are 7 A and 10 A respectively. If NA = 6.02 1023, find molecular weight of virus.
- 15.4 kg/mol
- 1.54 104 kg/mol
- 3.08 104 kg/mol
- 3.08 103 kg/mol
Answer
(1) 15.4 kg/mol
Solution
r = 7 A = 7 10-8 cm, l = 10 A = 10-7 cm
Volume of one particle, V = r2 l
= 227 (7 10-8)2 10-7
= 1.54 10-21 cc
Mass of one particle = volumespecific volume = 1.54 10-216.02 10-2 g
∴ Molecular weight = 1.54 10-216.02 10-2 6.02 1023
= 1.54 104 g/mol
= 15.4 kg/mol
Answer: (1)
AIPMT 2001