Crude oil pumped out of the ground may be accompanied by the formation of water, a solution that contains high concentrations of NaCl and other salts. If a boiling point of a sample of the formation of water is 2.0 Celsius above the boiling point of pure water, what is the molality of particles in the sample?
delta T = i*Kb*m
i for NaCl is 2. I assume the problem expects you to use i for NaCl since "other" salts are not listed.
To determine the molality of particles in the sample, we need to calculate the change in boiling point using the formula:
ΔTb = kbp * molality
Where:
ΔTb = change in boiling point
kbp = boiling point elevation constant (for water, it is approximately 0.52°C/m)
molality = molality of particles in the sample
Given that the boiling point of the sample is 2.0°C above the boiling point of pure water, we have:
ΔTb = 2.0°C
kbp = 0.52°C/m
Substituting these values into the formula, we can solve for the molality:
2.0°C = 0.52°C/m * molality
Simplifying the equation, we find:
molality = 2.0°C / 0.52°C/m
molality ≈ 3.85 mol/kg
Therefore, the molality of particles in the sample is approximately 3.85 mol/kg.
To determine the molality of particles in the sample, we need to use the boiling point elevation formula. Boiling point elevation is a colligative property, meaning that it depends on the concentration of particles in a solution rather than the specific type of particles.
The equation for boiling point elevation is:
∆T = K_b * m
Where:
∆T is the boiling point elevation (given as 2.0 Celsius in the question)
K_b is the molal boiling point elevation constant (which depends on the solvent, in this case water)
m is the molality of particles in the solution (what we're trying to find)
We need to find the molal boiling point elevation constant (K_b) for water. This value is a constant specific to the solvent and is typically provided. For water, it is approximately 0.512 °C/m.
Now, let's substitute the given values into the equation:
2.0 °C = (0.512 °C/m) * m
We can solve this equation to find the value of m (molality). Rearranging the equation:
2.0 = 0.512m
Dividing both sides of the equation by 0.512:
m = 2.0 / 0.512
Calculating this value:
m ≈ 3.91 mol/kg
Therefore, the molality of particles in the sample is approximately 3.91 mol/kg.
I had this on my homework and got the correct answer.
delta T = i * m * Kb
delta T -- 2 * 2.0 (the difference in temperature)
i -- it is 2 for NaCl
Kb -- is the van't hoff factor of water; which is 0.52 degress C/m
Just solve for m