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

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: