Verified Solution • 2025–26 Session moderate

QAIPM-2001-Ch-2857: NEET 2001 Chemistry - Class 11 AIPM-2001-Ch-2857 Solution: Specific volume of cylindrical virus particle is \(6.02…

Specific volume of cylindrical virus particle is

6.02\times10^{-2}\ \text{cc/g}

whose radius and length are 7

\text{\AA}

and 10

\text{\AA}

respectively. If

N_A=6.02\times10^{23}

, find molecular weight of virus.

Multiple Choice Options

A 15.4 kg/mol
B $1.54\times10^4\ \text{kg/mol}$
C $3.08\times10^4\ \text{kg/mol}$
D $3.08\times10^3\ \text{kg/mol}$

Step-by-Step Verified Solution

Specific volume ( vol. of 1 g) cylindrical virus particle = 6.02 _ 10 -2 cc/g Radius of virus, r = 7 � = 7 _ 10 -8 cm Volume of virus = � r 2 l = 22 7 _ ( 7 _ 10 _ 8 ) _ 10 _ 10 _ 8 = 154 _ 10 -23 cc wt. of one virus particle = V o l u m e S p e c i f i c V o l u m e _ 1.54 _ 10 _ 23 6.02 _ 10 _ 2 g _ Molecular wt of virus = wt. of N A particle = 1.54 _ 10 _ 23 6.02 _ 10 _ 2 _ 6.02 _ 10 _ 23 g / m o l = 15400 g/mol = 15.4 kg/mol

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