If you add 2g of NaCl to 23mL of water, what is the boiling point?
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2 g/molar mass Salt = 58.4425 mole of salt
delta T = Kb*m
delta T = 0.512 x 0.23 = 0.113 or about 113 C.
I tried to model this problem like you did, but I don't know if I did it correctly.
delta T = Kim
delta T = the change in boiling (or freezing) point (in degrees Celsius)
K= boiling point elevation (or depression) constant for the solvent
i = number of ions into which the solute dissociates
m = molality (moles solute / kg solvent)
Yes, delta T = iKb*molality is correct.
In this case, sucrose doesn't ionize; therefore, i is 1.
I suppose I should have said, "In the other problem, sucrose doesn't ionize and i = 1." In the problem with NaCl, i = 2.
To calculate the boiling point elevation, you need to use the formula ΔT = Kb * m. Here's how you can solve it step by step:
1. Determine the molar mass of NaCl: The molar mass of NaCl is approximately 58.4425 g/mol.
2. Calculate the moles of NaCl: Divide the mass of NaCl (2g) by its molar mass. 2g / 58.4425 g/mol = 0.0342 mol.
3. Determine the molality (m) of the solution: Molality is defined as the moles of solute (NaCl) divided by the mass of the solvent (water) in kilograms. The mass of water is 23 mL, which is equivalent to 0.023 kg.
molality (m) = moles of solute (NaCl) / mass of solvent (water in kg)
= 0.0342 mol / 0.023 kg
= 1.487 mol/kg
4. Retrieve the boiling point elevation constant (Kb) for water: The Kb value for water is 0.512 °C/m.
5. Calculate the boiling point elevation (ΔT): Multiply Kb by the molality (m) to find the boiling point elevation.
ΔT = Kb * m
= 0.512 °C/m * 1.487 mol/kg
≈ 0.761344 °C
6. Convert the boiling point elevation from Celsius to Fahrenheit (optional): If you prefer Fahrenheit, you can convert the value. To convert Celsius to Fahrenheit, use the formula F = (C * 9/5) + 32.
ΔT in Fahrenheit = (0.761344 °C * 9/5) + 32
≈ 33.37 °F
Therefore, by adding 2g of NaCl to 23 mL of water, the boiling point of the solution would increase by approximately 0.761344 °C (33.37 °F).