The number of ions present in 2 L of a solut | vedclass

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(C) The molarity of the solution is $1.6 \ M$ and the volume is $2 \ L$. Number of moles of $K_4[Fe(CN)_6] = 	ext{Molarity} 	imes 	ext{Volume} = 1.

Vedclass · Accepted Answer

$9.6 	imes 10^{24}$

Vedclass · Answer

(C) The molarity of the solution is $1.6 \ M$ and the volume is $2 \ L$. Number of moles of $K_4[Fe(CN)_6] = 	ext{Molarity} 	imes 	ext{Volume} = 1.6 	imes 2 = 3.2 \ mol$. Each formula unit 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 number of ions $= 	ext{Number of moles} 	imes N_A 	imes 5$. Total number of ions $= 3.2 	imes 6.022 	imes 10^{23} 	imes 5 = 16 	imes 6.022 	imes 10^{23} \approx 9.635 	imes 10^{24}$.

The number of ions present in $2 \ L$ of a solution of $1.6 \ M \ K_4[Fe(CN)_6]$ is:

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