Sp. vol of cylinderical virus particle is 6.02\times10^{-2} cc/gm. Whose radius and length are 7 Å & 10 Å respectively. If \N_\A=6.02\times10^23. Find mol. wt. of virus :-

Correct Answer is : 1.54 kg/mol

AIPMT 2001 Physics and Chemistry Paper

Question : 76 of 100

Marks: +1, -0

Sp. vol of cylinderical virus particle is 6.02×10−2 cc/gm. Whose radius and length are 7 Å & 10 Å respectively. If NA=6.02×1023. Find mol. wt. of virus :-

1.54 kg/mol
1.54×104 kg/mol
3.08×104 kg/mol
3.08×103 kg/mol

Validate

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