Answer :
Answer:
For a: The molar mass of the gene fragment is 19182 g/mol
For b: The freezing point depression of this solution is [tex]3.41\times 10^{-5}^oC[/tex]
Explanation:
- For a:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
Or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 0.340 torr
i = Van't hoff factor = 1 (for non-electrolytes)
Given mass of sample = 11.3 mg = 0.0113 g (Conversion factor: 1 g = 1000 mg)
Molar mass of gene fragment = ?
Volume of solution = 32.2 mL
R = Gas constant = [tex]62.364\text{ L.torr }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[273+25]=298K[/tex]
Putting values in above equation, we get:
[tex]0.340torr=1\times \frac{0.0113\times 1000}{\text{Molar mass of gene fragment}\times 32.2}\times 62.364\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of gene fragment}=\frac{1\times 0.0113\times 1000\times 62.364\times 298}{0.340\times 32.2}=19182g/mol[/tex]
Hence, the molar mass of the gene fragment is 19182 g/mol
- For b:
To calculate the mass of solution, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 0.997 g/mL
Volume of solution = 32.2 mL
Putting values in above equation, we get:
[tex]0.997g/mL=\frac{\text{Mass of solution}}{32.2mL}\\\\\text{Mass of solution}=(0.997g/mL\times 32.2mL)=32.1034g[/tex]
Mass of solvent = [32.1034 - 0.0113] = 32.0921 g
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
To calculate the depression in freezing point, we use the equation:
[tex]\Delta T_f=iK_fm[/tex]
Or,
[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 1 (For non-electrolytes)
[tex]K_f[/tex] = molal freezing point elevation constant = 1.86°C/m
[tex]m_{solute}[/tex] = Given mass of solute (gene fragment) = 0.0113 g
[tex]M_{solute}[/tex] = Molar mass of solute (gene fragment) = 19182 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = 32.0921 g
Putting values in above equation, we get:
[tex]\Delta T_f=1\times 1.86^oC/m\times \frac{0.0113\times 1000}{19182g/mol\times 32.0921}\\\\\Delta T_f=3.41\times 10^{-5}^oC[/tex]
Hence, the freezing point depression of this solution is [tex]3.41\times 10^{-5}^oC[/tex]