Calculate boiling point and freezing point of this solution which contains 5.85g of C6H12O6 in 255g H2O. pleasee
i've read formulas to use ro solve this problem but i don't know what delta T means. please tell me what formula to use and what delta T means thaaaank youu
As far as I know, Delta T means the difference in temperature. T final- T initial.
First find molar mass of C6H12O6:
I found it to be 180.1566 g/mol
now find the ratio of moles
5.85/180.1566 = .0325mol
Molality of C6H12O6:
(.0325*1000)/255 = .127
note: the value 1000 is from the density of water which is 1.00g/mL
So freezing point:
delta_T = -k_f * m
k_f = 1.86 for water
so delta_T = -1.86*.127 + 0 = -.236 Celcius
For boiling:
delta_T = k_b*m
where k_b =.512
so
.512*.127 + 100 = 100.065 Celcius
Some check my work
THANK YOU SO MUCH!
please explain more about how to find the molality??
oh i got it thank you!
To calculate the boiling point and freezing point of a solution, we need to determine the change in boiling point and freezing point caused by the presence of a solute in the solvent. We can use colligative properties to do this.
Colligative properties are properties of solutions that depend on the number of solute particles but not their nature. Two important colligative properties are boiling point elevation and freezing point depression.
1. Boiling Point Elevation:
The boiling point elevation (ΔTb) is given by the formula:
ΔTb = Kb * m * i
where Kb is the molal boiling point elevation constant for the solvent, m is the molality of the solution, and i is the van't Hoff factor.
2. Freezing Point Depression:
The freezing point depression (ΔTf) is given by the formula:
ΔTf = Kf * m * i
where Kf is the molal freezing point depression constant for the solvent, m is the molality of the solution, and i is the van't Hoff factor.
Step 1: Calculate the molality (m) of the solution.
Molality (m) is defined as the moles of solute per kilogram of solvent.
Given:
Mass of C6H12O6 = 5.85 g
Molar mass of C6H12O6 = 180.16 g/mol
Mass of H2O = 255 g
Molar mass of H2O = 18.01 g/mol
Moles of C6H12O6 = mass / molar mass = 5.85 g / 180.16 g/mol
Moles of H2O = mass / molar mass = 255 g / 18.01 g/mol
Molality (m) = moles of solute / mass of solvent (in kg)
m = (moles of C6H12O6 / mass of H2O) * 1000
Step 2: Determine the van't Hoff factor (i).
The van't Hoff factor (i) represents the number of particles formed when a solute dissolves in a solvent. For glucose (C6H12O6), it is 1 because it does not dissociate into ions.
Step 3: Look up the molal boiling point elevation constant (Kb) and molal freezing point depression constant (Kf) for water.
Kb for water = 0.512 °C/m
Kf for water = 1.86 °C/m
Step 4: Calculate the boiling point elevation and freezing point depression.
ΔTb = Kb * m * i
ΔTf = Kf * m * i
Step 5: Calculate the boiling point and freezing point of the solution.
Boiling point of solution = boiling point of pure solvent + ΔTb
Freezing point of solution = freezing point of pure solvent - ΔTf
Note: The normal boiling point of pure water is 100°C, and the normal freezing point is 0°C.
Now you can apply these steps to calculate the boiling point and freezing point of the given solution.