Which has the highest boiling point .33 m NH3 or .10 m Na2SO4
1 answer:
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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<u>Answer:</u> The empirical and molecular formula for the given organic compound is and
<u>Explanation:</u>
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
<u>For calculating the mass of carbon:</u>
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 13.00 g of carbon dioxide, of carbon will be contained.
<u>For calculating the mass of hydrogen:</u>
In 18 g of water, 2 g of hydrogen is contained.
So, in 2.662 g of water, of hydrogen will be contained.
Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.
For Carbon =
For Hydrogen =
For Oxygen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1 : 1 : 2
Hence, the empirical formula for the given compound is
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
We are given:
Mass of molecular formula = 90.04 g/mol
Mass of empirical formula = 45 g/mol
Putting values in above equation, we get:
Multiplying this valency by the subscript of every element of empirical formula, we get:
Hence, the empirical and molecular formula for the given organic compound is and