If ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ ionises as ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{O}} \to 2{{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{SO}}_4^2$, then total number of ions produced by ${\rm{0}}{\rm{.1}}\;{\rm{M}}\;{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ will be

JEE CHEMISTRY MCQ: SOME BASIC CONCEPTS OF CHEMISTRY

1 Easy

1 of 1

Application +4 | -1

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

ionises as

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + 2{\mathrm{H}_2}\mathrm{O} \to 2{\mathrm{H}_3}{\mathrm{O}^ + } + \mathrm{SO}_4^2

, then total number of ions produced by

\mathrm{0}\mathrm{.1}\;\mathrm{M}\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

will be

View answer and explanation

Correct option: D

1.8 \times {10^{23}}

Option D (

1.8 \times {10^{23}}

) is correct because 0.1 M

H_2SO_4

produces 0.3 M ions, which translates to

1.8 \times {10^{23}}

ions.