What is the boiling point of each solution?
a).50mol glucose in 1000g H2O
b)1.50mol NaCl in 1000g H2O
a. a. (T-100°C)= 0.512°C /molal x 0.50 mol/ 1.00 kg
T=100°C + 0.256°C = 100.256°C
0.50mol NaCl in1000gH2o
To determine the boiling point of a solution, we need to consider the concept of boiling point elevation. The boiling point of a solution is higher than that of the pure solvent due to the presence of solute particles, which disrupt the normal boiling process.
To calculate the boiling point elevation, we can use the equation:
ΔTb = Kb * m
Where:
ΔTb = boiling point elevation
Kb = molal boiling point constant (specific to each solvent)
m = molality of the solution (moles of solute per kilogram of solvent)
For water (H2O), the molal boiling point constant (Kb) is approximately 0.512 °C/m.
Now, let's calculate the boiling point of each solution:
a) 50 mol of glucose in 1000 g of water (H2O):
First, we need to convert the mass of water to kilograms:
Mass of water = 1000 g = 1 kg
Next, we calculate the molality (m) of the solution:
m = moles of solute / mass of solvent (in kg)
m = 50 mol / 1 kg = 50 mol/kg
Using the boiling point elevation equation:
ΔTb = Kb * m
ΔTb = 0.512 °C/m * 50 mol/kg = 25.6 °C
So, the boiling point of this solution is increased by 25.6 °C compared to pure water.
b) 1.50 mol of NaCl in 1000 g of water (H2O):
Again, we should convert the mass of water to kilograms:
Mass of water = 1000 g = 1 kg
Next, calculate the molality (m) of the solution:
m = moles of solute / mass of solvent (in kg)
m = 1.50 mol / 1 kg = 1.50 mol/kg
Using the boiling point elevation equation:
ΔTb = Kb * m
ΔTb = 0.512 °C/m * 1.50 mol/kg = 0.768 °C
Therefore, the boiling point of this solution is increased by 0.768 °C compared to pure water.
molality = moles/kg
Solve for molality.
delta T = Kb*molality
Solve for delta T.
Add to normal boiling point.