Question

She dissolves a 10.0mg sample in enough water to make 30.0mL of solution. The osmotic pressure of the solution is 0.340torr at 25C. a). What is the molar mass of the gene fragment? b). If the solution density is 0.997g/mL, what is the freezing point for this aqueous solution?

Answer 1

(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
`-3.413*10^(-5)` C

Step-by-step explanation:

The osmotic pressure (π) is given by the following equation:

`\pi =cRT`

`c`= Concentration of solution

R = universal gas constant = 62.364
`(L* torr)/(mol* K)`

T = temperature

Weight of solute = w = 10.0 mg

Let the molecular weight of the solute be m g/mol.

Concentration =
`c=(n)/(V)\\ n=(w)/(m)\\ n=(10*10^(-3))/(m)\\c=(10*10^(-3))/(m*30*10^(-3))M`

`m=(RT)/(3\pi)`

m = 18220.071g/mol

Therefore, the molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

`\Delta T_(f)=K_(f)m`

m is the molality of the solution.

m =
`1.835*10^(-5)` mol/kg

`\Delta T_(f)=1.86* m`

`\Delta T_(f)` =
`3.413*10^(-5)` C

The freezing point for the aqueous solution is
`-3.413*10^(-5)` C