Question

✦100 POINT QUESTION✦

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✯SUBJECT - CHEMISTRY✯



Specific volume of cylindrical virus particle is 6.02 * 10^-2 cc/gm whose radius and length are 7Å and 10Å respectively.If NA=6.02 *10^23 find molecular weight of virus



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Answer 1

i think this is better ans and you better understand ok
Specific volume (vol. of 1 gm) of
cylindrical virus particle = 6.02x10-2
c.c/gRadius of the virus (r) = 7A = 7
× 10-8 cmLength of the virus (l) =
10A = 10 × 10-8 cmNA (Avagadro’s
number) = 6.023 × 1023Volume of
virus = πr2l= 3.14 × (7x10-8)2 × 10 ×
10-8 cm= 154 × 10-23 c.c.Therefore,
weight of 1 virus particle = Volume /
Specific volume= 154 × 10-23/6.02 ×
10-2 gMolecular weight of virus =
Weight of Sodium (Na) particles= (154
× 10-23/6.02 × 10-2) × 6.023 × 1023=
15400 gm/mole= 15.4 kg/mole

Answer 2

Hey mate here is your answer

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Specific volume (vol. of 1 gm) of

cylindrical virus particle = 6.02x10-2

c.c/gRadius of the virus (r) = 7A = 7

× 10-8 cmLength of the virus (l) =

10A = 10 × 10-8 cmNA (Avagadro’s

number) = 6.023 × 1023Volume of

virus = πr2l= 3.14 × (7x10-8)2 × 10 ×

10-8 cm= 154 × 10-23 c.c.Therefore,

weight of 1 virus particle = Volume /

Specific volume= 154 × 10-23/6.02 ×

10-2 gMolecular weight of virus =

Weight of Sodium (Na) particles= (154

× 10-23/6.02 × 10-2) × 6.023 × 1023=

15400 gm/mole= 15.4 kg/mole

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