(A) . $16 \ g \ CH_{4} = 1 \ mol \ CH_{4}$. Total electrons in $1 \ molecule \ CH_{4} = 6 + 4 = 10$. Total electrons in $1 \ mol = 10 \times 6.02 \tim

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(A) . $16 \ g \ CH_{4} = 1 \ mol \ CH_{4}$. Total electrons in $1 \ molecule \ CH_{4} = 6 + 4 = 10$. Total electrons in $1 \ mol = 10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$. Thus,$A-II$.$B$. $1 \ g \ H_{2} = 0.5 \ mol \ H_{2}$. Volume at $STP$ = $0.5 \times 22.4 \ L = 11.2 \ L$. Thus,$B-IV$.$C$. $1 \ mol \ N_{2} = 1 \times 28 \ g = 28 \ g$. Thus,$C-I$.$D$. $0.5 \ mol \ SO_{2} = 0.5 \times 64 \ g = 32 \ g$. Thus,$D-III$.Therefore,the correct match is $A-II, B-IV, C-I, D-III$.

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

Difficult

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A$. $16 \ g \ CH_{4(g)}$ $I$. Weighs $28 \ g$

$B$. $1 \ g \ H_{2(g)}$ $II$. $60.2 \times 10^{23}$ electrons

$C$. $1 \ mole \ N_{2(g)}$ $III$. Weighs $32 \ g$

$D$. $0.5 \ mol \ SO_{2(g)}$ $IV$. Occupies $11.2 \ L$ volume at $STP$

Choose the correct answer from the options given below:

JEE MAIN 2023 All JEE MAIN

A
$A-II, B-IV, C-I, D-III$
B
$A-II, B-III, C-IV, D-I$
C
$A-II, B-IV, C-III, D-I$
D
$A-I, B-III, C-II, D-IV$

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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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JEE MAIN 2023 Paper All JEE MAIN years →

In which of the following reactions is no gas evolved on the product side? (Temperature is maintained below the boiling point of liquids formed)

Difficult

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Match the scientist's name in List-$A$ with their statement in List-$B$.

List-$A$ List-$B$

$(1)$ Roald Hoffmann $(A)$ The proportion of the mass of elements in any compound is fixed.

$(2)$ Lavoisier $(B)$ The smallest indivisible particle of matter is the atom.

$(3)$ Joseph Proust $(C)$ Chemistry is not the science of $100$ elements.

$(4)$ Dalton $(D)$ Matter can neither be created nor destroyed.

Easy

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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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Aluminum chloride ionizes in aqueous solution because ........

Medium

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$A$ mixture of sodium oxide $(Na_2O)$ and calcium oxide $(CaO)$ is dissolved in water and saturated with excess carbon dioxide $(CO_2)$ gas. The resulting solution is $......$ It contains $.........$ :

EasyAP EAMCET 2003

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List-$I$	List-$II$
$A$. $16 \ g \ CH_{4(g)}$	$I$. Weighs $28 \ g$
$B$. $1 \ g \ H_{2(g)}$	$II$. $60.2 \times 10^{23}$ electrons
$C$. $1 \ mole \ N_{2(g)}$	$III$. Weighs $32 \ g$
$D$. $0.5 \ mol \ SO_{2(g)}$	$IV$. Occupies $11.2 \ L$ volume at $STP$

List-$A$	List-$B$
$(1)$ Roald Hoffmann	$(A)$ The proportion of the mass of elements in any compound is fixed.
$(2)$ Lavoisier	$(B)$ The smallest indivisible particle of matter is the atom.
$(3)$ Joseph Proust	$(C)$ Chemistry is not the science of $100$ elements.
$(4)$ Dalton	$(D)$ Matter can neither be created nor destroyed.

Similar Questions

In which of the following reactions is no gas evolved on the product side? (Temperature is maintained below the boiling point of liquids formed)

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:

Aluminum chloride ionizes in aqueous solution because ........

$A$ mixture of sodium oxide $(Na_2O)$ and calcium oxide $(CaO)$ is dissolved in water and saturated with excess carbon dioxide $(CO_2)$ gas. The resulting solution is $......$ It contains $.........$ :

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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: