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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
States of Matter

NTA Abhyas 2020 · Chemistry · States of Matter

A mixture of C H _ 4 and C _ 2 H _ 2 occupied a certain volume at a...

A mixture of CH4C{H}_{4} and C2H2{C}_{2}{H}_{2} occupied a certain volume at a total pressure equal to 63 torr. The same gas mixture was burnt to CO2C{O}_{2} and H2O(l).CO2(g){H}_{2}O ( \text{l} ).C{O}_{2} ( g ) alone was collected in the same volume and at the same temperature, the pressure was found to be 69 torr.
What was the mole fraction of CH4C{H}_{4} in the original gas mixture?

Options

  1. A1921\dfrac{19}{21}
  2. B1920\dfrac{19}{20}
  3. C1718\dfrac{17}{18}
  4. D1516\dfrac{15}{16}

Correct answer

A. 1921\dfrac{19}{21}

Step-by-step solution

Let no. of moles of \text{C}{\text{H}_{4} present = n1{n}_{1} mole
Let no. of moles of C2H2{C}_{2} {H}_{2} present = n2{n}_{2} mole
\mathop{{\text{CH}_{4}}_{{n}_{1}} {\text{+2O}_{2} \to \mathop{{\text{CO}_{2}}_{{n}_{1}} {\text{+2H}_{2} \text{O}
\mathop{{C}_{2} {H}_{2}}_{{n}_{2}} \text{+}\dfrac{5}{2} {O}_{2} \to \text{2}\mathop{{\text{CO}_{2}}_{{\text{2n}_{2}} {\text{+H}_{2} \text{O}
Totalno . ofmolesatinitial = {n}_{1} + {n}_{2}<br/>Totalno.ofmolesatfinal=n1+2n2<br />Totalno . ofmolesatfinal = {n}_{1} + 2{n}_{2}
At constant V & T
P1P2=No.ofmolesatinitialNo.ofmolesatfinal\dfrac{{P}_{1}}{{P}_{2}} = \dfrac{No . ofmolesatinitial}{No . ofmolesatfinal}
6369=n1+n2n1+2n2⇒2123=n1+n2n1+2n2\dfrac{63}{69} = \dfrac{{n}_{1} + {n}_{2}}{{n}_{1} + 2{n}_{2}} ⇒ \dfrac{21}{23} = \dfrac{{n}_{1} + {n}_{2}}{{n}_{1} + 2{n}_{2}}
⇒n2n1=219⇒n1+n2n1=2119⇒ \dfrac{{n}_{2}}{{n}_{1}} = \dfrac{2}{19} ⇒ \dfrac{{n}_{1} + {n}_{2}}{{n}_{1}} = \dfrac{21}{19}
∴n1n1+n2=1921∴ \dfrac{{n}_{1}}{{n}_{1} + {n}_{2}} = \dfrac{19}{21}
Alternative solutions
{CH4(g)+2O2(g)→CO2(g)+2H2O(l)(63−x)Torr(63−x)Torr\begin{cases}{CH}_{4 \left(\text{g}\right)} & + & 2 {O}_{2 \left(\text{g}\right)} & \to & {CO}_{2 \left(\text{g}\right)} & + & 2 {H}_{2} \text{O} \left(\text{l}\right) \\ \left(6 3 - \text{x}\right) Torr & & & & \left(6 3 - \text{x}\right) \text{Torr} & & \end{cases}
{C2H2(g)+52O2(g)→2CO2(g)+H2O(l)(x)Torr(2x)Torr\begin{cases}{C}_{2} {H}_{2 \left(\text{g}\right)} & + & \dfrac{5}{2} {O}_{2 \left(\text{g}\right)} & \to & 2 {CO}_{2 \left(\text{g}\right)} & + & {H}_{2} \text{O} \left(\text{l}\right) \\ \left(\text{x}\right) Torr & & & & \left(2 \text{x}\right) \text{Torr} & & \end{cases}
∑PCO2=63−x+2x=63+x=69 Torr{∑}_{\text{P} {CO}_{2}} = 6 3 - \text{x} + 2 \text{x} = 6 3 + \text{x} = \text{69 Torr}
∴x=6 Torr∴ \text{x} = \text{6 Torr}
XCH4=63−663=5763=1921{X}_{\text{CH} 4} = \dfrac{6 3 - 6}{6 3} = \dfrac{5 7}{6 3} = \dfrac{1 9}{2 1}
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