Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation, $2{{\rm{H}}_2}{{\rm{O}}_2}\left( {aq} \right) \to 2{{\rm{H}}_2}{\rm{O}}\left( l \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$ under conditions where one mole of gas occupies 24 ${\rm{d}}{{\rm{m}}^3},100{\rm{c}}{{\rm{m}}^3}$ of $XM$ solution of ${{\rm{H}}_2}{{\rm{O}}_2}$ produces 3 ${\rm{d}}{{\rm{m}}^3}$ of ${{\rm{O}}_2}$. Thus, $X$ is :

JEE CHEMISTRY MCQ: SOME BASIC CONCEPTS OF CHEMISTRY

1 Medium

1 of 1

Application +4 | -1

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation,

2{\mathrm{H}_2}{\mathrm{O}_2}\left( {aq} \right) \to 2{\mathrm{H}_2}\mathrm{O}\left( l \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)

under conditions where one mole of gas occupies 24

\mathrm{d}{\mathrm{m}^3},100\mathrm{c}{\mathrm{m}^3}

XM

solution of

{\mathrm{H}_2}{\mathrm{O}_2}

produces 3

\mathrm{d}{\mathrm{m}^3}

{\mathrm{O}_2}

. Thus,

X

is :

View answer and explanation

Correct option: A

2.5

Option A (2.5) is correct because the molarity X is calculated based on the volume of oxygen produced from the decomposition of hydrogen peroxide.