Question

A biochemical engineer isolates a bacterial gene fragment and dissolves an 11.3 mg sample of the material in enough water to make 32.2 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25°C.

(a) What is the molar mass of the gene fragment?
(b) If the solution density is 0.997 g/mL, how large is the freezing point depression for this solution (Kf of water = 1.86 °C/m)?

Answer 1

For a: The molar mass of the gene fragment is 19182 g/mol

For b: The freezing point depression of this solution is
`3.41* 10^(-5)^oC`

Step-by-step explanation:

• For a:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

`\pi=iMRT`

Or,

`\pi=i* \frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}* RT`

where,

`\pi` = 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 =
`62.364\text{ L.torr }mol^(-1)K^(-1)`

T = temperature of the solution =
`25^oC=[273+25]=298K`

Putting values in above equation, we get:

`0.340torr=1* \frac{0.0113* 1000}{\text{Molar mass of gene fragment}* 32.2}* 62.364\text{ L. torr }mol^(-1)K^(-1)* 298K\\\\\text{Molar mass of gene fragment}=(1* 0.0113* 1000* 62.364* 298)/(0.340* 32.2)=19182g/mol`

Hence, the molar mass of the gene fragment is 19182 g/mol

• For b:

To calculate the mass of solution, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of solution = 0.997 g/mL

Volume of solution = 32.2 mL

Putting values in above equation, we get:

`0.997g/mL=\frac{\text{Mass of solution}}{32.2mL}\\\\\text{Mass of solution}=(0.997g/mL* 32.2mL)=32.1034g`

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:

`\Delta T_f=iK_fm`

Or,

`\Delta T_f=i* K_f* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}`

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

`K_f` = molal freezing point elevation constant = 1.86°C/m

`m_(solute)` = Given mass of solute (gene fragment) = 0.0113 g

`M_(solute)` = Molar mass of solute (gene fragment) = 19182 g/mol

`W_(solvent)` = Mass of solvent (water) = 32.0921 g

Putting values in above equation, we get:

`\Delta T_f=1* 1.86^oC/m* (0.0113* 1000)/(19182g/mol* 32.0921)\\\\\Delta T_f=3.41* 10^(-5)^oC`

Hence, the freezing point depression of this solution is
`3.41* 10^(-5)^oC`