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Bacteria

17. Specific ...

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17. Specific volume of cylindrical virus particle is 6.02 into 10 to the power minus 2 C C /gram whose radius and length are 7 angstromand 10 angstrom if Avogadro'snumber is 6.02 into 10 to the power 23 find molecular weight of virus

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Q. Specific volume of cylindrical virus particle is 6.02

\times 10^{- 2}

cc/gm whose radius and length are 7

˚ A

˚ A

respectively. If

N_{A} = 6.02 \times 10^{23}

, find the molecular weight of virus.

Q. Specific volume of cylindrical virus particles is

6.02 \times 10^{- 2}

cc/gm whose radius and length are 7

˚ A

˚ A

, respectively. If

N_{A} = 6.02 \times 10^{23}

find molecular weight of virus.

Q. Specific volume of cylindrical virus particle is

6.02 \times 10^{- 2} \frac{c c}{g}

, whose radius and length arc

7 \circ A

respectively. If

N_{A} = 6.02 \times 10^{23},

find molecular weight of virus.

Q. A plant virus is found to consist of uniform cylindrical particles of

150

angstrom in diameter and

5000

angstom long. The specific volume of virus is

0.75 {c m}^{3} g^{- 1}

. If the virus is considered to be a single particle, its molecular weight is :
(Given that volume of cylinder is

{π r}^{2} l

)

Q. A plant virus is found to consist of uniform cylindrical particle of

150

Angstrom in diameter and

5000

Angstrom long. The specific volume of the virus is

0.75 m L g^{- 1}

. if the virus is considered to be a single particle, find its molar mass.

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