Match List-I with List-II. List-I List | vedclass

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Question

(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 	imes 6.02 	im

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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 $STP$

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