Answer:
(a)The molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol
(b)The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C
Explanation:
The osmotic pressure (π) is given by the following equation:
[tex]\pi =cRT[/tex]
[tex]c[/tex]= Concentration of solution
R = universal gas constant = 62.364 [tex]\frac{L\times torr}{mol\times K}[/tex]
T = temperature
Weight of solute = w = 10.0 mg
Let the molecular weight of the solute be m g/mol.
Concentration = [tex]c=\frac{n}{V}\\ n=\frac{w}{m}\\ n=\frac{10\times10^{-3}}{m}\\c=\frac{10\times10^{-3}}{m\times30\times10^{-3}}M[/tex]
[tex]m=\frac{RT}{3\pi}[/tex]
m = 18220.071g/mol
Therefore, the molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol
[tex]\Delta T_{f}=K_{f}m[/tex]
m is the molality of the solution.
m = [tex]1.835\times10^{-5}[/tex] mol/kg
[tex]\Delta T_{f}=1.86\times m[/tex]
[tex]\Delta T_{f}[/tex] = [tex]3.413\times10^{-5}[/tex] C
The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C