calculate the boiling point of 25.5g of C7H11NO7S in 1.00x100g H20
delta T = Kb*m
Find m = molality. mols C7H11NO7S = grams/molar mass = ?
Then m = mols/kg solvent
Substitute Kb and m and find delta T and add to 100 C.
Post your work if you get stuck.
To calculate the boiling point of a solution, you need to determine the molality (moles of solute per kilogram of solvent) of the solute and water.
First, let's calculate the number of moles of C7H11NO7S (solute):
1. Determine the molar mass of C7H11NO7S:
C: 12.01 g/mol * 7 = 84.07 g/mol
H: 1.01 g/mol * 11 = 11.11 g/mol
N: 14.01 g/mol * 1 = 14.01 g/mol
O: 16.00 g/mol * 7 = 112.00 g/mol
S: 32.07 g/mol * 1 = 32.07 g/mol
Total molar mass = 84.07 + 11.11 + 14.01 + 112.00 + 32.07 = 253.26 g/mol
2. Convert the mass of C7H11NO7S to moles:
Moles of C7H11NO7S = 25.5 g / 253.26 g/mol = 0.1007 mol
Next, let's calculate the molality of the solute in the water:
3. Determine the mass of water (H2O) in kilograms:
Mass of water = 1.00 x 100 g = 100 g = 0.100 kg
4. Calculate the molality (moles of solute per kilogram of solvent):
Molality (m) = moles of solute / mass of solvent (in kg)
Molality (m) = 0.1007 mol / 0.100 kg = 1.007 mol/kg
Now, we can use the boiling point elevation equation to calculate the boiling point elevation (ΔTb):
5. Use the equation:
ΔTb = Kb * m
where:
ΔTb = boiling point elevation (change in boiling point)
Kb = molal boiling point elevation constant (a constant specific to the solvent; for water it is approximately 0.52 °C/m)
m = molality of the solution
For water, Kb = 0.52 °C/m
ΔTb = 0.52 °C/m * 1.007 mol/kg = 0.52364 °C
Finally, calculate the boiling point of the solution:
6. Boiling point of the solution = boiling point of pure water + ΔTb
Boiling point of pure water = 100°C (at standard atmospheric pressure)
Boiling point of the solution = 100°C + 0.52364 °C = 100.52364°C
Therefore, the boiling point of a solution containing 25.5g of C7H11NO7S in 100g of water is approximately 100.52°C.