10 , mL of concentrated HCl is diluted to 1 , | vedclass

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(C) Let the normality of the diluted $HCl$ solution be $N_1$ and its volume be $V_1 = 20 \, mL$.For $NaOH$,$N_2 = 0.1 \, N$ and $V_2 = 25 \, mL$.Using

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$12.5$

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(C) Let the normality of the diluted $HCl$ solution be $N_1$ and its volume be $V_1 = 20 \, mL$.For $NaOH$,$N_2 = 0.1 \, N$ and $V_2 = 25 \, mL$.Using the titration formula $N_1 V_1 = N_2 V_2$:$N_1 	imes 20 = 0.1 	imes 25$$N_1 = \frac{0.1 	imes 25}{20} = 0.125 \, N$.This $N_1$ is the normality of the diluted solution $(1 \, L)$.To find the normality of the original $10 \, mL$ concentrated $HCl$,we use the dilution equation $N_{conc} V_{conc} = N_{dil} V_{dil}$:$N_{conc} 	imes 10 \, mL = 0.125 \, N 	imes 1000 \, mL$$N_{conc} = \frac{0.125 	imes 1000}{10} = 12.5 \, N$.

$10 \, mL$ of concentrated $HCl$ is diluted to $1 \, L$. If $20 \, mL$ of this diluted solution is completely neutralized by $25 \, mL$ of $0.1 \, N$ sodium hydroxide solution,then the normality of the original concentrated hydrochloric acid is:

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