A colloidal solution of gold nanoparticles is prepared by reduction of $HAuCl_4$. Which of the following methods is MOST EFFECTIVE for purifying this solution from excess reactants and small ionic byproducts?

Reactants react in the equal number of …….. to give products.

How many gram of ${{\rm{I}}_2}$ are present in a solution which requires 40 mL, of 0.11 N ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ to react with it, ${{\rm{S}}_2}{\rm{O}}_3^{2 - } + {{\rm{I}}_2} o {{\rm{S}}_4}{\rm{O}}_6^{2 - } + 2{{\rm{I}}^ - }?$

For the reaction, $2{\rm{F}}{{\rm{e}}^{3 + }} + {\rm{S}}{{\rm{n}}^{2 + }} o 2{\rm{F}}{{\rm{e}}^{2 + }} + {\rm{S}}{{\rm{n}}^{4 + }}$ The normality of ${\rm{SnC}}{{\rm{l}}_2}$ (mol.wt. = 189.7) solution prepared by dissolving 47.5 g in acid solution and diluting with ${{\rm{H}}_2}{\rm{O}}$ to a total of 2.25 litre is :

In the following change, $3{\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} o {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}$. If the atomic weight of iron is 56, then its equivalent weight will be :

In the standardization of ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ using ${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ by iodometry, the equivalent weight of ${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ is :

In the following change, $3{\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} o {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}$. If the atomic weight of iron is 56, then its equivalent weight will be :

What is the normality of a ${\rm{KMn}}{{\rm{O}}_4}$ solution to be used as an oxidant in acid medium, which contain 15.8 g of the compound in 100 mL of solution? Mol. wt. of ${\rm{KMn}}{{\rm{O}}_4}$ is 158 :

What mass of ${\rm{Mn}}{{\rm{O}}_2}$ is reduced by 35 mL of 0.16 $N$ oxalic acid in acidic solution? The skeleton equation is, ${\rm{Mn}}{{\rm{O}}_2} + {{\rm{H}}^ + } + {{\rm{H}}_2}{{\rm{C}}_2}{{\rm{O}}_4} o {\rm{C}}{{\rm{O}}_2} + {{\rm{H}}_2}{\rm{O}} + {\rm{M}}{{\rm{n}}^{2 + }}:$

How many g of ${\rm{KMn}}{{\rm{O}}_4}$ are needed to prepare 3.75 litre of 0.850 $N$ solution if ${\rm{KMn}}{{\rm{O}}_4}$ is reduced as, ${\rm{MnO}}_4^ - + 8{{\rm{H}}^ + } + 5{\rm{e}} o {\rm{M}}{{\rm{n}}^{2 + }} + 4{{\rm{H}}_2}{\rm{O}}$

In the reaction, ${\rm{VO}} + {\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3} o {\rm{FeO}} + {{\rm{V}}_2}{{\rm{O}}_5}{\rm{.}}$. The eq.wt. of ${{\rm{V}}_2}{{\rm{O}}_5}$ is equal to its :