Calculate the freezing point and boiling point of 12.0g of glucose (C6H12O6) in 50.0g H20
Molar mass-180
12.0\180=
is the molar mass 6.67 or is it 6.67*10^-2
whats the kg solvent
The problem tell you 50.0 g H2O, which is 0.050 kg.
To calculate the freezing point and boiling point of a solution, we need to use the concept of molality and the molal freezing point depression and boiling point elevation constants for water.
Here's how we can solve this problem step by step:
Step 1: Calculate the number of moles of glucose (C6H12O6):
To find the number of moles of glucose, divide the given mass by the molar mass of glucose.
The molar mass of glucose (C6H12O6) is:
(6 * atomic mass of carbon) + (12 * atomic mass of hydrogen) + (6 * atomic mass of oxygen)
So,
molar mass of carbon = 12.01 g/mol
molar mass of hydrogen = 1.008 g/mol
molar mass of oxygen = 16.00 g/mol
molar mass of glucose (C6H12O6) = (6 * 12.01) + (12 * 1.008) + (6 * 16.00) = 180.18 g/mol
Now, divide the given mass of glucose (12.0 g) by the molar mass (180.18 g/mol) to find the number of moles:
Number of moles of glucose = mass of glucose / molar mass of glucose
Number of moles of glucose = 12.0 g / 180.18 g/mol
Step 2: Calculate the molality of the solution:
To find the molality (m) of the solution, divide the number of moles of solute by the mass of the solvent (water).
Number of moles of solute (glucose) = 12.0 g / 180.18 g/mol (calculated in step 1)
Mass of solvent (water) = 50.0 g
Molality (m) = moles of solute / mass of solvent
Molality (m) = (12.0 g / 180.18 g/mol) / 50.0 g
Step 3: Calculate the freezing point depression constant:
The freezing point depression constant (Kf) is a property of the solvent. For water, the value is 1.86 °C/m.
Step 4: Calculate the freezing point depression:
The freezing point depression (∆Tf) is calculated using the formula:
∆Tf = Kf * m
Substitute the values of Kf and m into the formula to find the freezing point depression (∆Tf) of the solution.
Step 5: Calculate the freezing point of the solution:
The freezing point of the pure solvent (water) is 0 °C. To find the freezing point of the solution, subtract the freezing point depression (∆Tf) from the freezing point of the pure solvent.
Freezing point (°C) = 0 °C - ∆Tf
Step 6: Calculate the boiling point elevation constant:
The boiling point elevation constant (Kb) is a property of the solvent. For water, the value is 0.52 °C/m.
Step 7: Calculate the boiling point elevation:
The boiling point elevation (∆Tb) is calculated using the formula:
∆Tb = Kb * m
Substitute the values of Kb and m into the formula to find the boiling point elevation (∆Tb) of the solution.
Step 8: Calculate the boiling point of the solution:
The boiling point of the pure solvent (water) is 100 °C. To find the boiling point of the solution, add the boiling point elevation (∆Tb) to the boiling point of the pure solvent.
Boiling point (°C) = 100 °C + ∆Tb
By following these steps, you can calculate the freezing point and boiling point of the given solution of glucose in water.
Neither. The molar mass = 180.
moles in 12.0 grams is
12.0/180 = 0.06667 and I would keep it like that until the end, then round the final answer.
freezing point.
delta T = Kf*molality.
Convert 12.0 g glucose to moles. Moles = grams/molar mass.
Then use m = moles/kg solvent to calculate molality, substitute that into the first equation to calculate delta T. Subtract from 0o to find the freezing point of the solution.
boiling point. Done essentially the same way.
delta T = Kb*m
Post your work if you get stuck.