How many grams of NaCl would need to be added to 5501 g of water to increase the boiling temperature of the solution by 1.500 °C? (Kb for water is 0.5100 °C/m)

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

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

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!