Question
A biomedical engineer is working in a lab that is isolating gene fragments. She dissolves a...
A biomedical engineer is working in a lab that is isolating gene fragments. 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. What is the molar mass of the gene fragment? If the solution density is 0.997g/mL, what is the freezing point for this aqueous solution?
Answers
Osmotic pressure
of a solution is calculated as follows:
Where C is the molar concentration
R is gas constant =
Hence, we can write
Hence, the concentration of gene fragment is
.
It is given that the solution volume is 30.0 mL = 0.030 L.
Hence, the number of moles of gene fragment in the 30.0 mL volume is
Hence, 10.0 mg sample of the gene fragment equals
. Hence, the molar mass can be calculated as
hence, the molar mass of the gene fragment is
(rounded to three significant figures).
Now, given that the density of the solution is 0.997 g/mL
The mass of the solution of volume 30.0 mL can be calculated as
Now, the presence of the gene fragment lowers the freezing point of the aqueous solution compared to that of pure water.
The change in freezing point is calculated as
Where i is the van't Hoff factor ( For a non-electrolyte like a gene fragment , i=1)
Kf is the molal freezing point depression constant of water =
m is the molality of solvent given in moles of solute per kg of solvent.
Now, we can calculate the molality of our solution as follows:
Moles of gene fragment =
Mass of solvent = Mass of solution - Mass of gene fragment = 29.91 g - 0.01 g = 29.90 g = 0.0299 kg
Hence,
Hence, we can calculate the depression in freezing point as follows
We know that the freezing point of pure water
= 0 oC
Hence, the freezing point of the solution can be calculated as
Note that practically the freezing point of the solution is almost close to zero as the moality of the solution is so low owing to the high molar mass of gene fragment.
We were unable to transcribe this imageII = CRT62 363 L torr K-1 mol-1T = 25°C = 25 + 273.15 K = 298.15 KII = Osmotic pressure = 0.340 torrII = CRT +0.340 torr = C x 62.363L torr K-1 mol-1 x 298.15 K 0.340 torr >C= 2 62.363 L torr K-1 mol-1 x 298 15 1.828 x 10 mol/L1.828 x 10-5 mol/L1.828 x 10-5 mol - x 0.030 L = 5.486 x 10-7 mol5.486 x 10-7 molM = 5.48 10.0 mg 5.486 x 10-7 mol ~18229 g/mol 1000 mg 1.82 x 104 g/mol1.82 x 104 g/molmass = volume x density = 30.0 mL x 0.9979, = 29.91 gAT, = ix K: xm1.86°C kg mol-15.486 x 10-7 mol5.486 x 10-7 mol molality = - " 0.0299 kg 1.835 x 10-5 mol/kgATrix Kxm=1x 1.86 °C kg mol-1 x 1.835 x 10-5 mol/kg AT = 3.41 x 10-5 °CWe were unable to transcribe this imageTy solution =T: - AT, = 0.0°C – 3.41 x 10-5 °C = -3.41 x 10-50
of a solution is calculated as follows:
.
. Hence, the molar mass can be calculated as
(rounded to three significant figures).
= 0 oC