
A mixture of ${{C}_{3}}{{H}_{8}}$ and $C{{H}_{4}}$ exerts a pressure of 320 mm of Hg at temperature T (K) in a V litre flask. On complete combustion, a gaseous mixture contains $C{{O}_{2}}$ only, and exerts a pressure of 448 mm of Hg under identical condition. Hence, mole fraction of ${{C}_{3}}{{H}_{8}}$ in the mixture is:
(A) 0.2
(B) 0.8
(C) 0.25
(D) 0.75
Answer
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Hint: There are five basic types of chemical reactions, which will be useful for predicting the products of unknown reactions. The types of chemical reactions are combination, decomposition, single-replacement, double-displacement reaction, and combustion reaction. This classification allows us to analyze the reactants and products of a given reaction. There is a way to express the relative amount of substances in a mixture with the mole fraction.
Complete step by step solution:
In the given combustion reaction,
Let V liter at temperature T (K), the pressure = 320 mm of Hg represents 1 mole
Then, V liter at temperature T (K), the pressure = 448 mm of Hg represents $\dfrac{448}{320}mol$ = 1.4 mol
The complete combustion reactions of a given mixture of ${{C}_{3}}{{H}_{8}}$ and $C{{H}_{4}}$are,
Let moles of ${{C}_{3}}{{H}_{8}}$ = x moles, then moles of $C{{O}_{2}}$ = 3x moles
${{C}_{3}}{{H}_{8}}(g)+5{{O}_{2}}\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)$
Moles of $C{{H}_{4}}$ = 1-x moles,
$C{{H}_{4}}(g)+5{{O}_{2}}\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)$
Moles of $C{{O}_{2}}$ produced = 3x+1-x = 1+2x
Moles of $C{{O}_{2}}$ produced = moles represents at V litre at temperature T (K) under pressure 448 mm of Hg
1+2x = 1.4
x=0.2
Therefore, mole fraction of ${{C}_{3}}{{H}_{8}}$ = $\dfrac{x}{1}$ =0.2
Hence, mole fraction of ${{C}_{3}}{{H}_{8}}$ in the mixture = 0.2
The correct answer is option A.
Note: The ratio of moles one substance in a mixture to the total number of moles of all substances is called mole fraction. The mole fraction is helpful to calculate the mixture of gases. This mole fraction represents the number of molecules of a particular component in a mixture and expresses the concentration of the solution.
Complete step by step solution:
In the given combustion reaction,
Let V liter at temperature T (K), the pressure = 320 mm of Hg represents 1 mole
Then, V liter at temperature T (K), the pressure = 448 mm of Hg represents $\dfrac{448}{320}mol$ = 1.4 mol
The complete combustion reactions of a given mixture of ${{C}_{3}}{{H}_{8}}$ and $C{{H}_{4}}$are,
Let moles of ${{C}_{3}}{{H}_{8}}$ = x moles, then moles of $C{{O}_{2}}$ = 3x moles
${{C}_{3}}{{H}_{8}}(g)+5{{O}_{2}}\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)$
Moles of $C{{H}_{4}}$ = 1-x moles,
$C{{H}_{4}}(g)+5{{O}_{2}}\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)$
Moles of $C{{O}_{2}}$ produced = 3x+1-x = 1+2x
Moles of $C{{O}_{2}}$ produced = moles represents at V litre at temperature T (K) under pressure 448 mm of Hg
1+2x = 1.4
x=0.2
Therefore, mole fraction of ${{C}_{3}}{{H}_{8}}$ = $\dfrac{x}{1}$ =0.2
Hence, mole fraction of ${{C}_{3}}{{H}_{8}}$ in the mixture = 0.2
The correct answer is option A.
Note: The ratio of moles one substance in a mixture to the total number of moles of all substances is called mole fraction. The mole fraction is helpful to calculate the mixture of gases. This mole fraction represents the number of molecules of a particular component in a mixture and expresses the concentration of the solution.
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