Find the increase in the boiling temperature of water in a solution obtained by dissolving 240 g of glucose (180.1 g/mol) in 2.5 kg of water. The boiling point elevation constant for water is 0.512 ºC /m.
To find the increase in the boiling temperature of water in a solution, we can use the equation:
ΔTb = Kb * m
where:
ΔTb is the increase in the boiling temperature,
Kb is the boiling point elevation constant for the solvent (water in this case),
m is the molality of the solute (glucose) in the solution.
First, we need to calculate the molality (m) of the glucose solution. Molality is defined as the number of moles of solute per kilogram of solvent.
The molar mass (M) of glucose is 180.1 g/mol, and the mass of glucose is 240 g. To convert the mass of glucose to moles, we divide it by the molar mass:
moles of glucose = mass of glucose / molar mass of glucose
= 240 g / 180.1 g/mol
≈ 1.332 mol
Next, we calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
= 1.332 mol / 2.5 kg
= 0.5328 mol/kg
Now, we have the molality (m) and the boiling point elevation constant (Kb), which is given as 0.512 ºC/m. We can plug these values into the equation to calculate the increase in boiling temperature (ΔTb):
ΔTb = Kb * m
= 0.512 ºC/m * 0.5328 mol/kg
≈ 0.2737 ºC
Therefore, the increase in the boiling temperature of the water in the glucose solution is approximately 0.2737 ºC.