The specific volume of a cylindrical macromolecule is measured to be $5.0 imes 10^{-2} \mathrm{~cm

Question

The specific volume of a cylindrical macromolecule is measured to be $5.0 	imes 10^{-2} \mathrm{~cm}^3\mathrm{/g}$. Its microscopic radius and length are found to be $10 \mathrm{~\AA}$ and $50 \mathrm{~\AA}$ respectively. Given $N_A = 6.022 	imes 10^{23} \mathrm{~mol}^{-1}$, the estimated molecular weight of this macromolecule is nearest to

Professor PC Thomas Classes · Answer

(a) : The physical volume of one single cylindrical particle is $V = \pi r^2 l = 3.14 	imes (10 	imes 10^{-8} \mathrm{~cm})^2 	imes (50 	imes 10^{-8} \mathrm{~cm}) = 1.57 	imes 10^{-20} \mathrm{~cm}^3$. The specific volume is the volume occupied by $1 \mathrm{~g}$ of the substance, so the mass of one single particle is $m = \frac{V}{	ext{Specific volume}} = \frac{1.57 	imes 10^{-20}}{5.0 	imes 10^{-2}} = 3.14 	imes 10^{-19} \mathrm{~g}$. The molecular weight (molar mass) is the mass of one mole of these particles: $MW = m 	imes N_A = 3.14 	imes 10^{-19} 	imes 6.022 	imes 10^{23} \approx 18.91 	imes 10^4 \mathrm{~g/mol} = 189.1 \mathrm{~kg/mol}$.