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Ainat [17]
3 years ago
12

A certain substance X has a normal boiling point of 117.8°C and a molal boiling point elevation constant =Kb·0.63°C·kgmol−1. A s

olution is prepared by dissolving some urea NH22CO in 900.g of X. This solution boils at 118.3°C. Calculate the mass of urea that was dissolved. Round your answer to 1 significant digit.
Chemistry
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

42.8 g was the mass of dissolved urea.

Explanation:

Boiling point elevation to solve this:

ΔT = Kb . m

where ΔT → Boiling T° of solution - Boiling T° of pure solvent

Let's replace the data given:

118.3°C - 117.8°C = 0.63°C/m . m

0.5°C / 0.63 m/°C = m → 0.793 mol/kg

Mass of solvent → 900 g → g to kg → 900 g . 1kg/1000 = 0.9 kg

Molality . kg = moles → 0.793 mol/kg . 0.9 kg = 0.714 mol

These are the moles of urea we used. Let's determine the mass dissolved.

0.714 mol . 60 g /1mol = 42.8 g

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Given 4.80g of ammonium carbonate, find:
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1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

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<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

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<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

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2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

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