Excess of $NaOH\, (aq)$ was added to $100\, mL$ of $FeCl_3\, (aq)$ resulting in $2.14\, g$ of $Fe(OH)_3$. The molarity of $FeCl_3\, (aq)$ is (Given molar mass of $Fe=56\, g\, mol^{-1}$ and molar mass of $Cl=35.5\, g\, mol^{-1}$) (in $, M$)

Volume of $\frac{N}{10}$ $H_2SO_4$ required to neutralize $10 \ mL$ of $5 \ N$ $Na_2CO_3$ is.....$mL$.

The yield of each step as actually carried out in the laboratory is given in the reaction scheme below. Calculate the overall percentage yield of the reaction.
$Me_3C-CO-CH_3$ $\xrightarrow{58\% (A)} Me_3C-CO-CH_2Br$ $\xrightarrow{54\% (B)} Me_3C-CH(OH)-CH_2Br$ $\xrightarrow{68\% (C)} Me_3C-CH(O)CH_2$

$6.023 \times 10^{22}$ molecules are present in $10 \ g$ of a substance $'x'.$ The molarity of a solution containing $5 \ g$ of substance $'x'$ in $2 \ L$ solution is.......... $\times 10^{-3}$

What will be the weight of lime $(CaO)$ obtained by heating $300 \ kg$ of $90\%$ pure limestone $(CaCO_3)$?

$250 \ mL$ of a sodium carbonate solution contains $2.65 \ g$ of $Na_2CO_3$. If $10 \ mL$ of this solution is diluted to $1 \ L$,what is the concentration of the resultant solution (in $M$)?