An element X forms a chloride with the formula XCl$_3$ and another chloride with the formula XCl$_5$. 0.1 moles of XCl$_3$ requires 60 mL of 1 N NaOH for complete hydrolysis, while 0.1 moles of XCl$_5$ requires 100 mL of 1 N NaOH. What is the equivalent weight of the element X in XCl$_3$ and XCl$_5$ respectively?

Cerric ammonium sulphate and potassium permanganate are used as oxidising agents in acidic medium for oxidation of ferrous ammonium sulphate to ferric sulphate. The ratio of number of moles of cerric ammonium sulphate required per mole of ferrous ammonium sulphate to the number of moles of ${\rm{KMn}}{{\rm{O}}_4}$ required per mole of ferrous ammonium sulphate, is

An element X forms a chloride with the formula XCl$_3$ and another chloride with the formula XCl$_5$. 0.1 moles of XCl$_3$ requires 60 mL of 1 N NaOH for complete hydrolysis, while 0.1 moles of XCl$_5$ requires 100 mL of 1 N NaOH. What is the equivalent weight of the element X in XCl$_3$ and XCl$_5$ respectively?

The eq. wt. of ${{\rm{I}}_2}$ in the change ${{\rm{I}}_2} o {\rm{IO}}_3^ -$ is :

56g of CO contains (NA = Avogadro’s No.)

The equivalent weight of iron in ${\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}$ would be :

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,

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

Mass ratio of $N{H_3}\,and{\rm{ }}C{O_2}$ for maximum product formation as per reaction : $2N{H_3} + C{O_2} o N{H_2}COON{H_4}$

The number of ${\rm{F}}{{\rm{e}}^{2 + }}$ ion oxidised by one mole of ${\rm{MnO}}_4^ -$ ions 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 :

When ${\rm{BrO}}_3^ -$ ion reacts with ${\rm{B}}{{\rm{r}}^ - }$ ion in acidic solution ${\rm{B}}{{\rm{r}}_2}$ is liberated. The equivalent weight of ${\rm{KBr}}{{\rm{O}}_3}$ is :