At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in

Question

At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height $(h)$ of the solution (density $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ) where $h$ is equal to 2.00 cm . If the concentration of the dilute solution of the macromolecule is $2.00 \mathrm{~g} \mathrm{dm}^{-3}$, the molar mass of the macromolecule is calculated to be $X 	imes 10^{4} \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $X$ is $\_\_\_\_$ .\ Use: Universal gas constant $(R)=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and acceleration due to gravity $(g)=10 \mathrm{~m} \mathrm{~s}^{-2}$

Professor PC Thomas Classes · Answer

(2.49): Height $(h)=2 \mathrm{~cm}$  $$ =2 	imes 10^{-2} \mathrm{~m} $$ $d=1 \mathrm{~g} / \mathrm{cm}^{3}=1000 \mathrm{~kg} / \mathrm{m}^{3}$\ $g=10 \mathrm{~ms}^{-2}$\ Osmotic pressure $(\pi)=h d g$\ $=2 	imes 10^{-2} 	imes 1000 	imes 10=200 \mathrm{Nm}^{-2}$\ Osmotic pressure can also be given as $\pi=C R T$\ $C=\frac{2.00}{	ext { Molar mass }} \mathrm{mol} / \mathrm{L}$\ $=\frac{2000}{	ext { Molar mass }} \mathrm{mol} / \mathrm{m}^{3}$\ $200=\frac{2000}{	ext { Molar mass }} 	imes 8.3 	imes 300$\ Molar mass $=\frac{2000 	imes 300 	imes 8.3}{200}$\ $=2.49 	imes 10^{4} \mathrm{~g} \mathrm{~mol}^{-1}$\