(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker: $NaOH + HCl \rightarrow NaCl + H_2O$ Initial moles of $NaOH = 2 \ M \times 0.01 \ L = 0

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(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker: $NaOH + HCl \rightarrow NaCl + H_2O$ Initial moles of $NaOH = 2 \ M \times 0.01 \ L = 0.02 \ mol$. Initial moles of $HCl = 1 \ M \times 0.02 \ L = 0.02 \ mol$. Since the moles are equal,the resulting solution is neutral ($NaCl$ solution). Step $2$: $10 \ mL$ of this neutral solution is added to a flask containing $2 \ moles$ of $HCl$. The total volume of the flask is made up to $100 \ mL$ $(0.1 \ L)$. Molarity of $HCl = \frac{\text{moles of } HCl}{\text{Volume of solution in } L} = \frac{2 \ mol}{0.1 \ L} = 20 \ M$.

Class 11 Chemistry Some Basic Concepts of Chemistry Mix Examples of Some Basic Concept of Chemistry

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$10 \ mL$ of $2 \ M$ $NaOH$ solution is added to $20 \ mL$ of $1 \ M$ $HCl$ solution kept in a beaker. Now,$10 \ mL$ of this mixture is poured into a volumetric flask of $100 \ mL$ containing $2 \ moles$ of $HCl$ and made the volume up to the mark with distilled water. The solution in this flask is $:$

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A
$0.2 \ M$ $NaCl$ solution
B
$20 \ M$ $HCl$ solution
C
$10 \ M$ $HCl$ solution
D
Neutral solution

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Class 11

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Some Basic Concepts of Chemistry

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Mix Examples of Some Basic Concept of Chemistry

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The volume of $HCl$,containing $73 \ g \ L^{-1}$,required to completely neutralize $NaOH$ obtained by reacting $0.69 \ g$ of metallic sodium with water,is $........ mL$. (Nearest Integer) (Given: molar masses of $Na, Cl, O, H$ are $23, 35.5, 16$ and $1 \ g \ mol^{-1}$ respectively)

DifficultJEE MAIN 2023

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State whether the following statements are true or false:
$(i)$ The molarity of a $0.6 \ N$ aqueous solution of ${H_3PO_3}$ is $1.2 \ M$.
$(ii)$ $\text{Empirical formula} = n \times \text{Molecular formula}$.
$(iii)$ In ${Cl_2O}$,the weight of chlorine combining with $8 \ g$ of oxygen is $35.5 \ g$.

Medium

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Match the following:

Column-$I$ Column-$II$

$A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$ $I$. $32 \ g$

$B$. $1.2 \times 10^{24}$ atoms of oxygen $II$. $66 \ g$

$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ $III$. $42 \ g$

$D$. $1.5 \ g$-molecule of $N_2$ $IV$. $16 \ g$

The correct match is:

Column-$I$	Column-$II$
$A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$	$I$. $32 \ g$
$B$. $1.2 \times 10^{24}$ atoms of oxygen	$II$. $66 \ g$
$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$	$III$. $42 \ g$
$D$. $1.5 \ g$-molecule of $N_2$	$IV$. $16 \ g$

Medium

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If the density of methanol is $0.793 \, kg \, L^{-1}$,what is its volume needed for making $2.5 \, L$ of its $0.25 \, M$ solution (in $, mL$)?

Difficult

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$20 \ mL$ of $2 \ M$ $NaOH$ solution is added to $400 \ mL$ of $0.5 \ M$ $NaOH$ solution. The final concentration of the solution is $.......... \times 10^{-2} \ M$. $(Nearest \ integer)$.

MediumJEE MAIN 2025

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The volume of $HCl$,containing $73 \ g \ L^{-1}$,required to completely neutralize $NaOH$ obtained by reacting $0.69 \ g$ of metallic sodium with water,is $........ mL$. (Nearest Integer) (Given: molar masses of $Na, Cl, O, H$ are $23, 35.5, 16$ and $1 \ g \ mol^{-1}$ respectively)

State whether the following statements are true or false:$(i)$ The molarity of a $0.6 \ N$ aqueous solution of ${H_3PO_3}$ is $1.2 \ M$.$(ii)$ $\text{Empirical formula} = n \times \text{Molecular formula}$.$(iii)$ In ${Cl_2O}$,the weight of chlorine combining with $8 \ g$ of oxygen is $35.5 \ g$.

Match the following: Column-$I$ Column-$II$ $A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$ $I$. $32 \ g$ $B$. $1.2 \times 10^{24}$ atoms of oxygen $II$. $66 \ g$ $C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ $III$. $42 \ g$ $D$. $1.5 \ g$-molecule of $N_2$ $IV$. $16 \ g$ The correct match is:

If the density of methanol is $0.793 \, kg \, L^{-1}$,what is its volume needed for making $2.5 \, L$ of its $0.25 \, M$ solution (in $, mL$)?

$20 \ mL$ of $2 \ M$ $NaOH$ solution is added to $400 \ mL$ of $0.5 \ M$ $NaOH$ solution. The final concentration of the solution is $.......... \times 10^{-2} \ M$. $(Nearest \ integer)$.

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Match the following:

Column-$I$ Column-$II$

$A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$ $I$. $32 \ g$

$B$. $1.2 \times 10^{24}$ atoms of oxygen $II$. $66 \ g$

$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ $III$. $42 \ g$

$D$. $1.5 \ g$-molecule of $N_2$ $IV$. $16 \ g$

The correct match is: