The number of sodium ions present in 0.5 mol | vedclass

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Question

(C) The chemical formula for sodium ferrocyanide is $Na_4[Fe(CN)_6]$.Each molecule of $Na_4[Fe(CN)_6]$ contains $4$ sodium ions $(Na^+)$.Therefore, $1

Vedclass · Accepted Answer

$12 	imes 10^{23}$

Vedclass · Answer

(C) The chemical formula for sodium ferrocyanide is $Na_4[Fe(CN)_6]$.Each molecule of $Na_4[Fe(CN)_6]$ contains $4$ sodium ions $(Na^+)$.Therefore, $1 \ mol$ of $Na_4[Fe(CN)_6]$ contains $4 \ mol$ of $Na^+$ ions.For $0.5 \ mol$ of $Na_4[Fe(CN)_6]$, the number of moles of $Na^+$ ions is $0.5 	imes 4 = 2 \ mol$.The number of $Na^+$ ions is calculated as $2 	imes N_A$, where $N_A \approx 6.022 	imes 10^{23} \ mol^{-1}$.Number of ions $= 2 	imes 6.022 	imes 10^{23} = 12.044 	imes 10^{23} \approx 12 	imes 10^{23}$.

List-$I$ (Mass of substance)	List-$II$ (Number of atoms)
$A$. $1.8 \text{ mg}$ water	$I$. $2 \times 10^{-4} \times N_{A}$
$B$. $9.8 \text{ mg}$ sulphuric acid	$II$. $1.5 \times 10^{-4} \times N_{A}$
$C$. $1.8 \text{ mg}$ carbon	$III$. $3 \times 10^{-4} \times N_{A}$
$D$. $5.85 \text{ mg}$ salt (NaCl)	$IV$. $7 \times 10^{-4} \times N_{A}$

The number of sodium ions present in $0.5 \ mol$ of sodium ferrocyanide is?

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