Answer:
1x10^–9 M
Explanation:
From the question given,
Concentration of hydronium ion, [H3O+] = 1x10^-5 M.
Concentration of Hydroxide ion, [OH-] =..?
The concentration of the hydroxide ion, [OH-] can be obtained as follow:
[H3O+] x [OH-] = 1x10^–14
1x10^-5 M x [OH-] = 1x10^–14
Divide both side by 1x10^-5
[OH-] = 1x10^–14 / 1x10^-5
[OH-] = 1x10^–9 M
<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams
<u>Explanation:</u>
We are given:
44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.
To calculate volume of a substance, we use the equation:

Density of solution = 1.343 g/mL
Mass of solution = 100 g
Putting values in above equation, we get:

To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:
In 77.46 mL of solution, mass of sulfuric acid present is 44 g
So, in 60 mL of solution, mass of sulfuric acid present will be = 
Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams
Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
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