The weight of oxalic acid that will be required to prepare a $1000 \ mL$ $(N / 20)$ solution is

  • A
    $126 / 100 \ g$
  • B
    $63 / 40 \ g$
  • C
    $63 / 20 \ g$
  • D
    $126 / 20 \ g$

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Why is it essential to wash the precipitate with water before estimating it quantitatively?

$10 \, mL$ of concentrated $HCl$ were diluted to $1 \, L$. $20 \, mL$ of this diluted solution required $25 \, mL$ of $0.1 \, N$ sodium hydroxide solution for complete neutralization. The normality of the concentrated hydrochloric acid will be:

$A$ $2.0 \, g$ sample containing $MnO_2$ is treated with $HCl$ liberating $Cl_2$. The $Cl_2$ gas is passed into a solution of $KI$ and $60.0 \, mL$ of $0.1 \, M \, Na_2S_2O_3$ is required to titrate the liberated iodine. The percentage of $MnO_2$ in the sample is $.....$ (Nearest integer).
[Atomic masses (in $u$): $Mn = 55, Cl = 35.5, O = 16, I = 127, Na = 23, K = 39, S = 32$]

The required amount of crystalline oxalic acid (equivalent weight = $63$) to prepare $250 \, mL$ of $N/10$ oxalic acid solution is: (in $, g$)

The weight of potassium dichromate (molecular weight $= 294$) required to prepare $250 \ mL$ of $0.04 \ N$ solution is (in $g$)

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