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jonny [76]
3 years ago
7

Calculate the mole fraction of each solute and solvent in the solution of 46.85 g of codeine, C 18 H 21 NO 3 , in 125.5 g of eth

anol, C 2 H 5 OH.
Chemistry
1 answer:
krok68 [10]3 years ago
5 0

mol fraction Codein : 0.054

mol fraction Ethanol : 0.946

<h3>Further explanation</h3>

Given

46.85 g of codeine

125.5 g of ethanol

Required

mol fraction

Solution

The mole fraction : the mole ratio of a substance to the mole of solution /mixture  

mol Codeine, C₁₈H₂₁NO₃(MW=299.4 g/mol) :

\tt mol=\dfrac{46.85}{299.4 g/mol}\\\\mol=0.1565

mol Ethanol, C₂H₅OH(MW=46.07 g/mol)

\tt mol=\dfrac{125.5}{46.07}=2.7241

mol of solution :

\tt mol~codein+mol~ethanol=0.1565+2.7241=2.8806

  • mol fraction Codein :

\tt \dfrac{0.1565}{2.8806}=0.054

mol fraction Ethanol :

\tt \dfrac{2.7241}{2.8806}=0.946

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Answer:

ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given) (exothermic)

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           3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g)  +  17H₂O(g)   +  10CO₂(g)

ΔH⁰(f):  3(-365.6)Kj     1(-301)Kj    14(0)Kj       3(0)Kj    17(-241.8)Kj    10(-393.5)Kj

            = -1096.8Kj     = -301Kj     = 0Kj         = 0Kj      = -4110.6Kj    = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

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∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)

ΔH°(rxn for 14.11g  NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given)

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