Compute the number of ions present in 5.85 , | vedclass

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Question

(A) The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \, g/mol$.Number of moles of $NaCl = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{5.85 \, g}{

Vedclass · Accepted Answer

$1.2042 	imes 10^{23}$

Vedclass · Answer

(A) The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \, g/mol$.Number of moles of $NaCl = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{5.85 \, g}{58.5 \, g/mol} = 0.1 \, mol$.Each formula unit of $NaCl$ dissociates into two ions: one $Na^+$ ion and one $Cl^-$ ion.Therefore,total moles of ions $= 0.1 \, mol 	imes 2 = 0.2 \, mol$.Number of ions $= 	ext{Moles of ions} 	imes 	ext{Avogadro constant}$.Number of ions $= 0.2 	imes 6.022 	imes 10^{23} = 1.2044 	imes 10^{23}$ (approximately $1.2042 	imes 10^{23}$ based on standard calculation).Thus,the correct option is $A$.

Compute the number of ions present in $5.85 \, g$ of sodium chloride.

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