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A plant virus is found to consists of uniform cylindrical particles of $150 A^{\circ}$ in diameter and $5000\;A^{\circ}$ long. The specific volume of the virus is $0.75\;cm^3/g$ If the virus is considered to be a single particle , find its molar mass.

$(a)\;7.09 \times 10^8\;g \\ (b)\;7.09 \times 10^7\;g \\(c)\;6.09 \times 10^7\;g \\(d)\;7.09 \times 10^6 \;g $
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1 Answer

Since the virus is cylinder so the volume of virus $=\pi r^2 l$
$\qquad= \large\frac{22}{7} \times \bigg( \large\frac{150}{2} \bigg)^2 $$ \times 10^{-6} \times 5000 \times 10^{-8}$
$\qquad= 0.884 \times 10^{-16}\;cm^3$
weight of one virus $= \large\frac{0.884 \times 10^{-16}}{0.75}$$g$
$\qquad= 1.178 \times 10^{-16}$
Hence gram molecular weight of virus = Weight of No virus.
= Weight of $6.023 \times 10^{23}\;virus$
$\qquad= 1.178 \times 10^{-16} \times 6.023 \times 10^{23}\;g$
$\qquad= 7.09 \times 10^{7}\;g$
Molecular weight of virus $=7.09 \times 10^7\;g$
Hence b is the correct answer.
answered Mar 28, 2014 by meena.p

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