JEE Main 2012 — Chemistry Physical Chemistry

2012mcqhard

When $\mathrm{CO}_{2(\mathrm{~g})}$ is passed over red hot coke it partially gets reduced to $\mathrm{CO}(g)$ . Upon passing $0.5 \mathrm{~L}$ of $\mathrm{CO}_2(g)$ over red hot coke, the total volume of the gases increased to $700 \mathrm{~mL}$ . The composition of the gaseous mixture at STP is

Official previous-year question

Held on 19 May 2012 · Verified 6 Jul 2026.

Options

A
$\mathrm{CO}_2=300 \mathrm{~mL} ; \mathrm{CO}=400 \mathrm{~mL}$
B
$\mathrm{CO}_2=0.0 \mathrm{~mL} ; \mathrm{CO}=700 \mathrm{~mL}$
C
$\mathrm{CO}_2=200 \mathrm{~mL} ; \mathrm{CO}=500 \mathrm{~mL}$
D
$\mathrm{CO}_2=350 \mathrm{~mL} ; \mathrm{CO}=350 \mathrm{~mL}$

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Solution

$\mathrm{CO}_2+\mathrm{C} \longrightarrow 2 \mathrm{CO}$ Stoichoimetry ratio is $1: 2$ AT STP, $P=1 \mathrm{~atm}, T=273 \mathrm{~K}, R=0.0821$ Initial moles of $\mathrm{CO}_2 ; n\left(\mathrm{CO}_2\right.$ initial $)=\frac{P V}{R T}$ $=\frac{1 \times 0.5}{0.0821 \times 273}=0.022 \mathrm{~mole}$ In final mixture no. of moles; $n\left(\mathrm{CO}_2 / \mathrm{CO}\right.$ mixture) $=\frac{1 \times 0.7}{0.0821 \times 273}=0.031$ Increase in volume is by $=0.031-0.022$ $=0.009$ mole of gas Final no. of moles of $\mathrm{CO}$ i.e. $n_{\text {(CO final) }}$ $\begin{aligned} & n_{\text {(CO final) }}=2 n_{\left(\mathrm{CO}_2 \text { initial }\right)}-n_{\left(\mathrm{CO}_2\right. \text { final) }} \\ & \quad=2\left(0.022-\mathrm{n}_{\left(\mathrm{CO}_2 \text { final }\right)}\right. \\ & \left.n_{\text {(CO final) }}=0.044-2 n_{\text {(CO }} \text { final }\right) \\ & \therefore \text { Now, } n_{\text {(CO final) }}+n_{\text {(CO final) }}=0.031 \\ & n_{\left(\mathrm{CO}_2\right. \text { final) }}=0.031-n_{\text {(CO final) })} \ldots \text { (iii) } \end{aligned}$ Substituting (ii) in eq. (i) $\begin{aligned} & n_{\text {(CO final) }}=0.044-2\left[0.031-n_{\text {(CO final) }}\right. \\ & n_{\text {(CO final) }}=0.044-0.062+2 n_{\text {(CO final })} \end{aligned}$ $n_{(\mathrm{CO} \text { final) }}=0.018 \mathrm{~mol} .$ Volume of $\mathrm{CO}=V=\frac{n R T}{P}=\frac{0.018 \times 0.0821 \times 273}{1}$ $=0.40$ Litre and volume of $\mathrm{CO}_2=0.7$ litre $-0.4$ litre $=0.3$ litre $\therefore \mathrm{CO}_2=300 \mathrm{~mL}, \mathrm{CO}=400 \mathrm{~mL}$

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Options

$\mathrm{CO}_2=300 \mathrm{~mL} ; \mathrm{CO}=400 \mathrm{~mL}$

$\mathrm{CO}_2=0.0 \mathrm{~mL} ; \mathrm{CO}=700 \mathrm{~mL}$

$\mathrm{CO}_2=200 \mathrm{~mL} ; \mathrm{CO}=500 \mathrm{~mL}$

$\mathrm{CO}_2=350 \mathrm{~mL} ; \mathrm{CO}=350 \mathrm{~mL}$

Solution

JEE Main 2012 — Chemistry Physical Chemistry

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JEE Main 2012 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry