Determine the boiling point elevation of H2O in each of the following solutions:
1)2.5 m solution of C6H12O6 (glucose)in H2O.
2) 3.20 g C6 H12 O6 in 1.00 kg H20
3) 20.0 C12 H22 O11 (sucrose) in 500.g H2O
delta T = Kb*m
yes but how would i find m?
#1 has m listed in the problem.
#2 and #3 has grams listed.
grams/molar mass = moles and
molality = mols/kg solvent.
i'm still confused
can you help me solve them?
Show your work on #2, following my instructions so I will know what your trouble is. I will help you through it.
To determine the boiling point elevation of a solution, you need to use the equation ΔTb = Kb * m * i, where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant for the solvent, m is the molality of the solute, and i is the van't Hoff factor.
1) For the 2.5 m solution of C6H12O6 (glucose) in H2O:
- First, calculate the molality (m) of the glucose solution. Molality is defined as the number of moles of solute per kilogram of solvent. We know the mass of glucose (C6H12O6) is 2.5 moles, and the mass of water (H2O) is 1000 g (assuming it's 1 kg). Thus, the molality is calculated as:
molality = moles of solute / mass of solvent (in kg)
molality = 2.5 moles / 1 kg = 2.5 m
- Next, you need to find the molal boiling point elevation constant (Kb) for water. The Kb value for water is approximately 0.512 °C/m.
- Finally, find the van't Hoff factor (i) for glucose. For glucose, the van't Hoff factor is 1 because it does not dissociate or ionize in water. So, i = 1.
Now you can calculate the boiling point elevation:
ΔTb = Kb * m * i
ΔTb = 0.512 °C/m * 2.5 m * 1
ΔTb = 1.28 °C
Therefore, the boiling point elevation of the 2.5 m glucose solution in water is 1.28 °C.
2) For the solution of 3.20 g C6 H12 O6 in 1.00 kg H2O:
- First, calculate the number of moles of C6 H12 O6. You can use its molar mass which is 180.16 g/mol.
moles of C6 H12 O6 = mass of solute / molar mass of solute
moles of C6 H12 O6 = 3.20 g / 180.16 g/mol = 0.01777 mol
- Next, calculate the molality (m) of the solution. We know the moles of solute and the mass of solvent (1.00 kg H2O).
molality = moles of solute / mass of solvent (in kg)
molality = 0.01777 mol / 1 kg = 0.01777 m
- Then, find the molal boiling point elevation constant (Kb) for water, which is still 0.512 °C/m.
- Finally, find the van't Hoff factor (i) for glucose. Since glucose does not dissociate or ionize, i = 1.
Now you can calculate the boiling point elevation:
ΔTb = Kb * m * i
ΔTb = 0.512 °C/m * 0.01777 m * 1
ΔTb ≈ 0.0091 °C
Therefore, the boiling point elevation of the solution containing 3.20 g C6 H12 O6 in 1.00 kg H2O is approximately 0.0091 °C.
3) For the solution of 20.0 g C12 H22 O11 (sucrose) in 500 g H2O:
- First, calculate the moles of C12 H22 O11. You can use its molar mass which is 342.30 g/mol.
moles of C12 H22 O11 = mass of solute / molar mass of solute
moles of C12 H22 O11 = 20.0 g / 342.30 g/mol = 0.0584 mol
- Next, calculate the molality (m) of the solution. We know the moles of solute and the mass of solvent (500 g H2O).
molality = moles of solute / mass of solvent (in kg)
molality = 0.0584 mol / 0.500 kg = 0.117 m
- Then, find the molal boiling point elevation constant (Kb) for water, which is still 0.512 °C/m.
- Finally, find the van't Hoff factor (i) for sucrose. Sucrose does not dissociate or ionize in water, so i = 1.
Now you can calculate the boiling point elevation:
ΔTb = Kb * m * i
ΔTb = 0.512 °C/m * 0.117 m * 1
ΔTb ≈ 0.060 °C
Therefore, the boiling point elevation of the solution containing 20.0 g C12 H22 O11 in 500 g H2O is approximately 0.060 °C.