The number of ions present in 2.0 L of a sol | vedclass

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Question

(A) The molarity of the solution is $0.8 \ M$ and the volume is $2.0 \ L$. Number of moles of $K_4[Fe(CN)_6] = 	ext{Molarity} 	imes 	ext{Volume} =

Vedclass · Accepted Answer

$4.8 	imes 10^{24}$

Vedclass · Answer

(A) The molarity of the solution is $0.8 \ M$ and the volume is $2.0 \ L$. Number of moles of $K_4[Fe(CN)_6] = 	ext{Molarity} 	imes 	ext{Volume} = 0.8 	imes 2.0 = 1.6 \ 	ext{mol}$. One mole of $K_4[Fe(CN)_6]$ dissociates as: $K_4[Fe(CN)_6] ightarrow 4K^+ + [Fe(CN)_6]^{4-}$. Total number of ions produced per formula unit = $4 + 1 = 5$ ions. Total moles of ions = $1.6 \ 	ext{mol} 	imes 5 = 8.0 \ 	ext{mol}$. Total number of ions = $	ext{Moles of ions} 	imes N_A = 8.0 	imes 6.022 	imes 10^{23} = 48.176 	imes 10^{23} = 4.8176 	imes 10^{24}$ ions. Rounding to the nearest option,the answer is $4.8 	imes 10^{24}$.

The number of ions present in $2.0 \ L$ of a solution of $0.8 \ M \ K_4[Fe(CN)_6]$ is

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