In a titration, 10.0 ml of 0.0750 M HCl(aq) is exactly neutralized by 30.0 ml of KOH(aq) of unknown concentration. What is the c
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D.
D.
Answer: 400 g of solute is present in the solution
Explanation:
A solution consists of solute and solvent. A solute is defined as the component that is present in a smaller proportion and solvent is defined as the component that is present in a larger proportion.
Saturation point of a solution is defined as the point after which the solution becomes saturated. This simply means no more solute can be added to the solution.
Mass of solvent = 800 g
At saturation point, 500 g of solvent can dissolve 250 g of solute
Applying unitary method:
800 g of solvent will dissolve = of solute
Hence, 400 g of solute is present in the solution.
Answer:
1. Silver acetate → 166.87 g/m, 2.50 moles, 417.2 g, 1.50×10²⁴ particles
2. Glucose → 180 g/mol, 1.8 moles, 324 g, 1.08×10²⁴ particles
3. Lead sulfide → 239.26 g/m, 0.522 moles, 125 g, 3.14×10²³ particles
4. Iron (III) Chloride → 162.2 g/m, 0.390 moles, 63.3 g, 2.35×10²³ particles
5. Aluminum sulfate → 342.14 g/m, 1.56 mol, 533.7 g, 9.39×10²³ particles
6. Caffeine → 194 g/m, 7.17 moles, 1392 g, 4.32×10²⁴ particles
13.1, 83.9 L of N₂
13.2, 7.59 L of C₂H₆
13.3 232.8 L of SO₃
Explanation:
- Silver acetate → AgCH₃COO
Molar mass Ag + 2 molar mass C + 3 molar mass H + 2 molar mass O ⇒ 107.87 g/m + 2 . 12 g/m + 3 . 1 g/m + 2 . 16 g/m = 166.87 g/m (molar mass)
If we have 2.50 moles , we have → 2.5 mol . 166.87 g/m = 417.2 g
1 mol of salt has 6.02×10²³ representative particles
2.5 moles of salt must have (2.5 . 6.02×10²³) / 1 = 1.50×10²⁴ particles
- Glucose → C₆H₁₂O₆
Molar mass → 180 g/mol
Let's convert the mass to moles → 324 g . 1 mol / 180 g = 1.8 moles
To determine number of particles → 1.8 mol . 6.02×10²³ particles / 1 mol =
1.08×10²⁴ particles
- PbS → Lead sulfide → Molar mass = Molar mass Pb + Molar mass S
207.2 g/m + 32.06 g/m = 239.26 g/m
Mass to moles → 125 g . 1 mol / 239.26 g = 0.522 moles
0.522 moles . 6.02×10²³ particles / 1 mol = 3.14×10²³ representative particles.
- FeCl₃ → Iron(III) chloride
Molar mass = Molar mass Fe + 3 Molar mass Cl
55.85 g/m + 3 . 35.45 g/m = 162.2 g/m
With the representative particles, we determine the moles.
2.35×10²³ particles . 1 mol / 6.02×10²³ particles = 0.390 moles
0.390 mol . 162.2 g / 1 mol = 63.3 g
- Aluminum sulfate → Al₂(SO₄)₃
Molar mass → 2 molar mass Al + 3 molar mass S + 12 molar mass O
2. 26.98 g/m + 3 . 32.06 g/m + 12 . 16 g/m = 342.14 g/m
We determine mass → 342.14 g /m . 1.56 mol = 533.7 g
1.56 moles . 6.02×10²³ particles / 1 mol = 9.39×10²³ particles
- Caffeine → C₈H₁₀N₄O₂
Molar mass → 8 . 12 g/m + 10 . 1 g/m + 4 . 14 g/m + 2. 16 g/m = 194 g/m
We determine the moles, by the representative particles:
4.32×10²⁴ particles . 1mol / 6.02×10²³ = 7.17 moles
We convert the moles to mass, 7.17 mol . 194g / 1 mol = 1392 g
Excercise 13:
1. P . V = n . R . T
V = n . R . T / P → 3.75 mol . 0.082L.atm /mol.K . 273K / 1 atm = 83.9L
2. P .V = n . R . T
V = n . R . T / P → 0.339 mol . 0.082L.atm /mol.K . 273K / 1 atm = 7.59L
3. We determine the moles of SO₃ → mass / molar mass
835 g / 80.06 g/m = 10.4 moles
V = n . R . T / P → 10.4 mol . 0.082L.atm /mol.K . 273K / 1 atm = 232.8L
