Match the following:

Column-I
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(B) . $5.6 \ L$ of $SO_{2(g)}$ at $STP$ = $(5.6 / 22.4) \ mol = 0.25 \ mol$. Mass = $0.25 	imes 64 \ g = 16 \ g$ $(IV)$.
$B$. $1.2 	imes 10^{24}$ a

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) . $5.6 \ L$ of $SO_{2(g)}$ at $STP$ = $(5.6 / 22.4) \ mol = 0.25 \ mol$. Mass = $0.25 	imes 64 \ g = 16 \ g$ $(IV)$.
$B$. $1.2 	imes 10^{24}$ atoms of oxygen. If we assume $O_2$ molecule,$1.2 	imes 10^{24} / 6.022 	imes 10^{23} \approx 2 \ mol$ of $O$ atoms. Mass = $2 	imes 16 \ g = 32 \ g$ $(I)$.
$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ = $(33.6 / 22.4) \ mol = 1.5 \ mol$. Mass = $1.5 	imes 44 \ g = 66 \ g$ $(II)$.
$D$. $1.5 \ g$-molecule of $N_2$ = $1.5 \ mol$ of $N_2$. Mass = $1.5 	imes 28 \ g = 42 \ g$ $(III)$.
Thus,the correct match is $A-IV, B-I, C-II, D-III$.

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 $N_2$	$IV$. $16 \ g$

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:

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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: