(C) Let the normality of the diluted $HCl$ solution be $N_1$. Using the neutralization formula $N_1V_1 = N_2V_2$ for $HCl$ and $NaOH$: $N_1 \times 20

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(C) Let the normality of the diluted $HCl$ solution be $N_1$. Using the neutralization formula $N_1V_1 = N_2V_2$ for $HCl$ and $NaOH$: $N_1 \times 20 \, mL = 0.1 \, N \times 25 \, mL$ $N_1 = \frac{0.1 \times 25}{20} = 0.125 \, N$. The dilution factor for the concentrated $HCl$ is $\frac{1000 \, mL}{10 \, mL} = 100$. Therefore,the normality of the concentrated $HCl$ is $0.125 \times 100 = 12.5 \, N$.

Class 12 Chemistry Principles Related to Practical Chemistry Volumetric Analysis

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$10 \, mL$ 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
$8$
B
$9.5$
C
$12.5$
D
$15$

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$10 \, mL$ 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:

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In the standardization of $Na_2S_2O_3$ using $K_2Cr_2O_7$ by iodometry,the equivalent weight of $K_2Cr_2O_7$ is

Assertion : $K_2Cr_2O_7$ is used as a primary standard in volumetric analysis.Reason : It has a good solubility in water.

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

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Assertion : $K_2Cr_2O_7$ is used as a primary standard in volumetric analysis.
Reason : It has a good solubility in water.