The molar ratio of ${H_2}{\rm{ }}and{\rm{ }}{O_2}$ gases in a mixture which contains these two gases in mass ratio of 1 : 4 will be

Number of ${{\rm{K}}^ + }$ ions and mole of ${{\rm{K}}^ + }$ ions present in 1 litre of $\frac{N}{5}{\rm{KMn}}{{\rm{O}}_4}$ acidified solution respectively are :

What volume of ${{\rm{O}}_2}$ measured at standard conditions will be formed by the action of 100 mL of 0.5$N{\rm{KMn}}{{\rm{O}}_4}$ on hydrogen peroxide in an acidic solution? The skeleton equation for the reaction is,

${\rm{KMn}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + {{\rm{H}}_2}{{\rm{O}}_2} o {\rm{KHS}}{{\rm{O}}_4} + {\rm{MnS}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{O}} + {{\rm{O}}_2}:$∶

Maximum number of atoms are present in

Equivalent mass of ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ in its reaction with ${{\rm{I}}_2}$ is equal to :

The number of equivalent per mole of ${{\rm{H}}_2}{\rm{S}}$ used in its oxidation to ${\rm{S}}{{\rm{O}}_2}$ is :

The equivalent weight of ${\rm{MnS}}{{\rm{O}}_4}$ is half of its molecular weight when it is converted to :

A mixture of $NaHCO_3$ and $Na_2CO_3$ is titrated against $0.5$ N $HCl$ using phenolphthalein as the indicator. $10$ mL of the mixture consumes $4$ mL of $HCl$ to reach the endpoint. Methyl orange is then added to the same solution, and a further $6$ mL of $HCl$ is required to reach the second endpoint. What is the equivalent weight of the mixture?

The equivalent weight of an element is 12. What is the weight of the element that combines with 8g of oxygen?

The eq. wt. of ${\rm{KMn}}{{\rm{O}}_4}$ in the reaction, ${\rm{MnO}}_4^ - + {\rm{M}}{{\rm{n}}^{2 + }} + {{\rm{H}}_2}{\rm{O}} o {\rm{Mn}}{{\rm{O}}_2} + {{\rm{H}}^ + }$ (unbalanced) is :

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation, $2{{\rm{H}}_2}{{\rm{O}}_2}\left( {aq} \right) o 2{{\rm{H}}_2}{\rm{O}}\left( l \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$ 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 :