For each of these solutions, determine the boiling point of the solution
0.238 M glucose (C6H12O6) in water (Assume the density of the solution is 1.00 g/mL.)
moles = grams/molar mass
Solve for moles.
molality = moles/kg solvent
Solve for molality
delta T = Kb*molality
Solve for delta T, then add to 100 C to determine the boiling point.
Thanks
To determine the boiling point of a solution, we need to consider the concept of boiling point elevation. Boiling point elevation occurs when a solute is dissolved in a solvent, such as water, resulting in an increase in the boiling point of the solution compared to the pure solvent.
In order to calculate the boiling point elevation, we can use the equation:
ΔTb = i * Kb * m
Where:
- ΔTb is the boiling point elevation
- i is the van't Hoff factor, which represents the number of particles formed when the solute dissolves
- Kb is the molal boiling point elevation constant, which is dependent on the solvent
- m is the molality of the solution, which is the moles of solute per kilogram of solvent
In this case, we have a 0.238 M (mol/L) solution of glucose (C6H12O6) in water. First, we need to convert the molarity to molality by considering the density of the solution.
To calculate the molality of the solution, we need to know the molar mass of glucose. The molar mass of glucose (C6H12O6) can be calculated by adding the atomic masses of carbon (12.01 g/mol), hydrogen (1.01 g/mol), and oxygen (16.00 g/mol), giving us a total molar mass of 180.18 g/mol.
Using the given molarity of 0.238 M and the molar mass of glucose, we can calculate the moles of glucose in 1 L of solution:
moles of glucose = molarity * volume of solution
moles of glucose = 0.238 mol/L * 1 L = 0.238 mol
Since the density of the solution is given as 1.00 g/mL (or 1.00 kg/L), the mass of the solvent (water) is also 1.00 kg.
Now, we can calculate the molality (m) of the solution:
molality (m) = moles of solute / mass of solvent (kg)
molality (m) = 0.238 mol / 1.00 kg = 0.238 mol/kg
Next, we need to determine the van't Hoff factor (i) for glucose in water. Since glucose is a non-electrolyte, it does not dissociate into ions upon dissolving. Therefore, the van't Hoff factor for glucose in water is i = 1.
Lastly, we need to find the molal boiling point elevation constant (Kb) for water. The value for Kb is 0.512 °C/m for water.
Now, we can substitute all the values into the boiling point elevation equation:
ΔTb = i * Kb * m
ΔTb = 1 * 0.512 °C/m * 0.238 mol/kg = 0.1224 °C
So, the boiling point of the solution is elevated by 0.1224 °C compared to the boiling point of pure water. To find the boiling point of the solution, we need to add the boiling point elevation to the boiling point of pure water at atmospheric pressure, which is 100 °C.
Boiling point of the solution = boiling point of pure water + ΔTb
Boiling point of the solution = 100 °C + 0.1224 °C = 100.1224 °C
Therefore, the boiling point of the solution containing 0.238 M glucose (C6H12O6) in water is approximately 100.1224 °C.