Question
Chemistry Question on Some basic concepts of chemistry
Sp. vol. of cylinderical virus particle is 6.02 × 10^-2 cc/gm. Whose radius and length are 7 Å & 10Å respectively. If NA = 6.02 × 10^23. Find mol. wt. of virus
A
1.54 kg/mol
B
1.54 × 10^4 kg/mol.
C
3.08 × 10^4 kg/mol.
D
3.08 × 10^3kg/mol.
Answer
1.54 kg/mol
Explanation
Solution
Sp. vol (vol. of 1gm) cylindrical virus particle = 6.02 × 10–2 cc/gm radius of virus r = 7Å = 7 × 10–8 cm length of virus = πr2l=22/7x (7 × 10^–8)^2 × 10 × 10^–8 = 154×10^–23cc wt. of one virus particle = Vol/Sp.vol. ⇒ 154x10^23/6.02x10^-2gm ∴ mol. wt. of virus = wt. of NA particles =154x10^-23/6.02x10^-2x6.02x10^+23gm/mol. = 15400 gm/mol = 15.4 kg/mol .