JEE Advance - Chemistry (1999 - No. 4)
A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single particle, find its molar mass.
7.09 × 10^5 g/mol
7.09 × 10^6 g/mol
7.09 × 10^7 g/mol
7.09 × 10^8 g/mol
7.09 × 10^9 g/mol
Explanation
Volume of a cylinder = $$\pi$$r2h
Radius of virus $$ = {{150} \over 2} = 75\mathop A\limits^o = 75 \times {10^{ - 8}}$$ cm.
$$\therefore$$ Volume of one virus
$$ = {{22} \over 7} \times {(75 \times {10^{ - 8}})^2} \times 5000 \times {10^{ - 8}}$$ cm3
$$ = 8.8397 \times {10^{ - 17}}$$ cm3
Mass of one virus
$$ = {{Volume} \over {Specific\,volume}} = {{8.8393 \times {{10}^{ - 17}}\,c{m^3}} \over {0.75\,c{m^3}\,{g^ - }}}$$
$$ = 1.178 \times {10^{ - 16}}\,g$$
Mol. wt. of virus
$$ = 1.178 \times {10^{ - 16}} \times 6.02 \times {10^{23}}\,g$$ mol$$-$$1
$$ = 7.09 \times {10^7}\,g$$ mol$$-$$1
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