In a mountainous location, the boiling point of pure water is found to be 95°C. How many grams of sodium chloride must be added to 9 kg of water to bring the boiling point back to 99°C? Assume that i = 2.
So what will my answer be in grams?
delta T =i*Kb*molality
Solve for molality. i = 2 for NaCl.
molality = moles/kg solvent.
Solve for moles.
moles = grams/molar mass.
Solve for grams.
To solve this problem, we need to use the equation for calculating the boiling point elevation:
ΔTb = i * K * m
Where:
ΔTb represents the change in boiling point
i is the van't Hoff factor (assuming it is 2 in this case for sodium chloride)
K is the molal boiling point elevation constant (which is a characteristic of the solvent, water in this case)
m is the molality of the solute (sodium chloride)
Given:
Initial boiling point (Tb1) = 95°C
Final boiling point (Tb2) = 99°C
Mass of water (mWater) = 9 kg
i = 2 (for sodium chloride)
First, we need to find the molal boiling point elevation constant (K) for water. The molal boiling point elevation constant for water is approximately 0.51°C·kg/mol.
Next, we need to calculate the change in boiling point (ΔTb) using the following formula:
ΔTb = Tb2 - Tb1
ΔTb = 99°C - 95°C = 4°C
Now, we can calculate the molality of the sodium chloride (m) using the following formula:
m = (ΔTb) / (i * K)
m = 4°C / (2 * 0.51°C·kg/mol) ≈ 3.92 mol/kg
Finally, to determine the mass of sodium chloride required, we can use the equation:
mass of sodium chloride = m * mWater
mass of sodium chloride = 3.92 mol/kg * 9 kg
mass of sodium chloride ≈ 35.28 kg
So, the answer will be approximately 35.28 kg of sodium chloride.