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Mama L [17]
3 years ago
7

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 2

5C. 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?
Chemistry
1 answer:
uranmaximum [27]3 years ago
8 0

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 -3.413\times10^{-5} C

Explanation:

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

\pi =cRT

c= Concentration of solution

R = universal gas constant = 62.364 \frac{L\times torr}{mol\times K}

T = temperature

Weight of solute = w = 10.0 mg

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

Concentration = c=\frac{n}{V}\\ n=\frac{w}{m}\\ n=\frac{10\times10^{-3}}{m}\\c=\frac{10\times10^{-3}}{m\times30\times10^{-3}}M

m=\frac{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\times10^{-5} mol/kg

\Delta T_{f}=1.86\times m

\Delta T_{f} = 3.413\times10^{-5} C

The freezing point for the aqueous solution is -3.413\times10^{-5} C

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