Answer:
476 g of NaCl would be need to be added
Explanation:
We apply the colligative property of Elevation of boiling point to solve this problem:
ΔT = Kb . m . i
ΔT refers to T° Eb of solution - T° Eb of pure solvent
Kb = Ebullioscopic constant
m = molality
i = Van't Hoff factor (indicates the ions which are dissolved)
NaCl → Na⁺ + Cl⁻ (for this case, i = 2)
Let's replace data:
1.5°C = 0.51 °C/ m . m . 2
m = 1.5°C / 0.51 m/°C . 2 = 1.48 mol/kg
To determine the moles of solute, we apply molality . mass of solvent (kg)
We convert the 5501 g to kg → 5501 g. 1 kg / 1000g = 5.501 kg
1.48 mol/kg . 5.501 kg = 8.14 moles
These are the moles of NaCl that must be added. Let's determine the mass → 8.14 mol . 58.45 g/mol = 476 g
Answer:
We have to add 472.5 grams of NaCl
Explanation:
Step 1: Data given
Molar mass NaCl = 58.44 g/mol
Mass of water = 5501 grams = 5.501 kg
Boiling point elevation = 1.500 °C
Boiling point elevation constant for water = 0.510 °C/m
Step 2: Calculate molality
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = 1.500 °C
⇒i = the van't Hoff factor for NaCl = 2 :(Nacl = Na+ + Cl-)
⇒Kb = the boiling point elevation constant = 0.510 °C /m
⇒m = molality = moles NaCl / mass water = TO BE DETERMINED
1.500 °C = 2* 0.510 °C/m * m
m = 1.47 molal
Step 3: Calculate moles NaCl
Moles NaCl = molality * mass water
Moles NaCl = 1.47 molal * 5.501 kg
Moles NaCl = 8.086 moles NaCl
Step 4: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 8.086 moles * 58.44 g/mol
Mass NaCl = 472.5 grams
We have to add 472.5 grams of NaCl
