Equivalent Concept

Vapour density of a metal chloride is 66. Its oxide contains 53% metal. The atomic weight of the metal is

Calculate the number of moles of $\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$ required to react with one mole of sulphite ion in an acidic solution.

$13\mathrm{g}$ sample of ${\mathrm{CaCO}}_3$ was dissolved in $0.5 \mathrm{L}$ of $0.1 \mathrm{M} \mathrm{HCl}$. The solution was made free from ${\mathrm{CO}}_2$ by boiling and the remaining solution required $\mathrm{x} \mathrm{mL}$ of $1 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ for complete neutralisation. Assuming no change in volume due to boiling, find the value of $\mathrm{x}$.

$0.177\mathrm{g}$ of a monobasic acid required $30\mathrm{ml} \mathrm{of} \mathrm{N}/10 \mathrm{NaOH}$ solution for complete neutralization. Its molecular weight will be

The valency factor of ammonium sulphate is

The valecy factor of Aluminium sulphate is

${\mathrm{KMnO}}_4$ reacts with ${\mathrm{H}}_2{\mathrm{O}}_2$ in an acidic medium. The number of moles of oxygen produced per mole of ${\mathrm{KMnO}}_4$ is;

If $6\times {10}^{-3} \mathrm{mole}$ of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ reacts completely with $9\times {10}^{-3}$ mole of ${\mathrm{X}}^{\mathrm{n}+}$ to give ${\mathrm{XO}}_3^{-} \text{and} {\mathrm{Cr}}^{3+}$, then, the value of $\mathrm{n}$ is

The equivalent weight of an element is $4$. Its chloride has a vapour density $59.25$. Then the valency of the element is

Arsenic forms two oxides, one of which contains $65.2\%$ and the other $75.7\%$ of the element. Hence, equivalent masses of arsenic are in the ratio :

What volume of $0.2 \mathrm{M} {\mathrm{KMnO}}_4$ is required to react with $1.58 \mathrm{g}$ of hypo solution $\left({\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\right)$ in acidic medium?

Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant? 

Volume ${\mathrm{V}}_1 \mathrm{mL}$ of $0.1\mathrm{M} {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is needed for complete oxidation of $0.678\mathrm{g} {\mathrm{N}}_2{\mathrm{H}}_4$ in acidic medium. The volume $\left(\mathrm{in} \mathrm{mL}\right)$ of $0.3 \mathrm{M} {\mathrm{KMnO}}_4$ needed for same oxidation in the acidic medium will be

In alkaline medium,${\mathrm{KMnO}}_4$ reacts as follows (Atomic weights $\mathrm{K}=39.09, \mathrm{Mn}=54.94, \mathrm{O}=16.00$)

$2{\mathrm{KMnO}}_4+2\mathrm{KOH}\to 2{\mathrm{K}}_2{\mathrm{MnO}}_4 +{\mathrm{H}}_2\mathrm{O}+\left[\mathrm{O}\right]$

Hence, its equivalent weight is

Equivalent weight of chlorine molecule in the equation is : $3{\mathrm{Cl}}_2+ 6\mathrm{NaOH} \to 5\mathrm{NaCl}+ {\mathrm{NaClO}}_3+3{\mathrm{H}}_2\mathrm{O}$

An aqueous solution of $6.3 \mathrm{g}$ of oxalic acid dihydrate is made up to $250 \mathrm{mL}$. The volume of $0.1 \mathrm{N} \mathrm{NaOH}$ required to completely neutralize $10 \mathrm{mL}$ of this solution is

In the conversion of ${\mathrm{NH}}_2\mathrm{OH}\to {\mathrm{N}}_2\mathrm{O}$, the equivalent weight of ${\mathrm{NH}}_2\mathrm{OH}$ is

($\mathrm{M}=$molecular weight of ${\mathrm{NH}}_2\mathrm{OH}$)