A biochemical engineer isolates a bacterial gene fragment and dissolves a 15.1 mg sample of the material in enough water to make 32.7 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)?
Convert 340 torr to atmospheres. torr/760 = atm
Then pi = MRT.
Substitute and solve for M = molarity.
Convert M to mols; M = mols/L. You have M and L, solve for mols.
Convert mols to molar mass; mols = grams/molar mass. You have mols and grams, solve for molar mass.
Can you follow up with b?
To answer these questions, we need to use the equations relating osmotic pressure and freezing point depression to calculate the molar mass and freezing point depression of the solution.
a) To find the molar mass of the gene fragment, we can use the equation for osmotic pressure:
Π = n/VRT
Where Π is the osmotic pressure, n is the number of moles of solute, V is the volume of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.
Rearranging this equation, we have:
n = Π * V * RT
First, we need to convert the osmotic pressure from torr to atm, since the ideal gas constant is in units of atm. We divide the osmotic pressure by the conversion factor of 760 torr/atm:
Π = 0.340 torr / 760 torr/atm = 0.000447 atm
Next, convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Substituting the values into the equation, we have:
n = 0.000447 atm * 0.0327 L * (0.0821 L·atm/(K·mol)) * 298.15 K
Simplifying, we find:
n = 0.00399 mol
Next, we can calculate the molar mass by dividing the mass of the gene fragment by the number of moles:
Molar Mass = Mass / Moles
Molar Mass = 15.1 mg / 0.00399 mol
Converting the mass to grams:
Molar Mass = 0.0151 g / 0.00399 mol
Molar Mass = 3.78 g/mol
Therefore, the molar mass of the gene fragment is approximately 3.78 g/mol.
b) To calculate the freezing point depression, we can use the equation:
ΔT = Kf * m
Where ΔT is the freezing point depression, Kf is the cryoscopic constant for water (1.86 °C/m), and m is the molality of the solution.
First, let's calculate the molality of the solution. Molality (m) is defined as moles of solute per kg of solvent.
We know that the mass of the solution is 15.1 mg dissolved in enough water to make 32.7 mL of solution.
To convert the mass of the solution to kg:
Mass of Solution = 0.0151 g + 32.7 mL * 0.997 g/mL
Mass of Solution = 48.4937 g
Mass of Solvent (water) = Mass of Solution - Mass of Solute
Mass of Solvent = 48.4937 g - 0.0151 g = 48.4786 g
Next, convert the mass of the solvent to kg:
Mass of Solvent = 48.4786 g / 1000 = 0.0484786 kg
Now, we can calculate the molality:
molality (m) = moles of solute / mass of solvent (kg)
molality (m) = 0.00399 mol / 0.0484786 kg
molality (m) = 0.0823 mol/kg
Finally, using the equation for freezing point depression, we can calculate ΔT:
ΔT = Kf * m
ΔT = 1.86 °C/m * 0.0823 mol/kg
ΔT = 0.153 °C
Therefore, the freezing point depression for this solution is approximately 0.153 °C.