MCQ
A gaseous hydrocarbon gives upon combustion $0.72\, g$ of water and $3.08\, g$ of $CO_2.$ The empirical formula of the hydrocarbon is :
  • A
    $C_2H_4$
  • B
    $C_3H_4$
  • C
    $C_6H_5$
  • $C_7H_8$

Answer

Correct option: D.
$C_7H_8$
d
$18 \mathrm{g} \mathrm{H}_{2} \mathrm{O}$ contain $2 \mathrm{g} \mathrm{H}$

$\therefore 0.72 \mathrm{g} \mathrm{H}_{2} \mathrm{O}$ contain $0.08 \mathrm{g} \mathrm{H}$

$44 \mathrm{g} \mathrm{CO}_{2}$ contain $12 \mathrm{g} \mathrm{C}$

$\therefore 3.08 \mathrm{g} \mathrm{CO}_{2}$ contain $0.84 \mathrm{g} \mathrm{C}$

$C: H=\frac{0.84}{12}: \frac{0.08}{1}=0.07: 0.08=7: 8$

$\therefore$ Empirical formula $=\mathrm{C}_{7} \mathrm{H}_{8}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The density of $2\; \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ is $1.28 \;\mathrm{g} / \mathrm{cm}^{3} .$ The molality of the solution is.....$m$ [Given that molecular mass of $\mathrm{NaOH}=40 \;\mathrm{g} \mathrm{mol}^{-1}$]
The bond having the highest bond energy is
Which of the following forms the most basic hydroxide?
You are given $500\, mL$ of $2\,N\, HCl$ and $500\, mL$ of $5\,N\, HCl$. What will be the maximum volume of $3\, M\, HCl$ that you can make from these two sotutions ? .............. $\mathrm{mL}$
Given below are two statements.

Statement $I:$ In the titration between strong acid and weak base methyl orange is suitable as an indicator.

Statement $II:$ For titration of acetic acid with $\mathrm{NaOH}$ phenolphthalein is not a suitable indicator.

In the light of the above statements, choose the most appropriate answer from the options given below:

The $pH$ of a solution obtained by mixing $50\,\,ml$ of $0.4\,\,N\,\,HCl$ and $50\,\,ml$ of $0.2\,\,N\,\,NaOH$ is
From the ground state electronic configuration of the elements given below, pick up the  one with highest value of second ionization energy
Acid precipitation is now regarded as a serious problem in some European and Asian countries. Its major cause or source is :
$HCl$ does not show peroxide effect since
$C{{H}_{3}}-CH=C{{H}_{2}}+\overset{\centerdot }{\mathop{X}}\,\to C{{H}_{3}}-\overset{\centerdot }{\mathop{C}}\,H-C{{H}_{2}}X$ $(I)$
$C{{H}_{3}}-\overset{\centerdot \,\,\,\,}{\mathop{CH}}\,-C{{H}_{2}}X+H-X\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}X+\overset{\centerdot }{\mathop{X}}\,$  $(II)$
The first ionization potentials in electron volts of nitrogen and oxygen atoms are respectively given by