Specific volume ( vol. of 1 g) cylindrical virus particle

= 6.02 $$ \times $$ 10^-2 cc/g

Radius of virus, r = 7 Å = 7 $$ \times $$ 10^-8 cm

Volume of virus = $$\pi {r^2}l$$

= $${{22} \over 7} \times (7 \times {10^{ - 8}}) \times 10 \times {10^{ - 8}}$$

= 154 $$ \times $$ 10^-23 cc

wt. of one virus particle = $${{Volume} \over {Specific\,\,Volume}}$$

$$ \Rightarrow $$ $${{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}}g$$

$$ \therefore $$ Molecular wt of virus = wt. of N_A particle

= $${{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}} \times 6.02 \times 10^{-23} \,\,g/mol$$

= 15400 g/mol = 15.4 kg/mol

$${{100} \over {0.5}} \times 78.4 = 1.568 \times {10^4}$$

Molarity = $${{1.17 \times 1000} \over {36.5 \times 1}} = 32.05$$

In A₃(BC₄)₂, (+2) $$ \times $$ 3 + 2[+5+4(-2)]

$$ \Rightarrow $$ +6 + 10 -16 = 0

$$ \therefore $$ Oxidation numbers of A, B, C are + 2, +5 and $$-$$2 respectively.