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Sagot :
Answer :-
[tex]\sf C) 1.8 \times 10^{23}[/tex]
Explanation:-
[tex]\red{ \underline { \boxed{ \sf{Moles\: of \: Ca(OH)_2 = \frac{mass\: of\: Ca(OH)_2}{molar\:mass \: of \: Ca(OH)_2}}}}}[/tex]
[tex]\begin{gathered}\\\implies\quad \sf Moles \:of \:Ca(OH)_2 = \frac{7.41}{74.1} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf Moles\: of\: Ca(OH)_2 = \frac{1}{10}\\\end{gathered}[/tex]
[tex]\sf Ca(OH)_2[/tex] dissociate into ions as-
[tex]\green{ \underline { \boxed{ \sf{Ca(OH)_2 \longrightarrow Ca^{2+} + 2OH^{-}}}}}[/tex]
[tex]\leadsto[/tex] 1 mole of [tex]\sf Ca(OH)_2[/tex] will dissociate into 1 mole of [tex]\sf Ca^{2+}[/tex] and 2 moles of [tex]\sf 2OH^{-}[/tex]
[tex]\leadsto[/tex]Similarly, 1\10 mole of [tex]\sf Ca(OH)_2[/tex] will dissociate into 1\10 mole of [tex]\sf Ca^{2+}[/tex] and 2\10 moles of [tex]\sf 2OH^{-}[/tex]
[tex]\\[/tex]
[tex]\maltese[/tex] Total number of ions = [tex]\sf Number\: of \:Ca^{2+} ion [/tex] +[tex]\sf Number\: of \:0H^{-} ion [/tex]
[tex]\begin{gathered}\\\implies\quad \sf \frac{1}{\cancel{10}} \times 6.022 \times \cancel{10^{23}}+ \frac{2}{ \cancel{10}} \times 6.022 \times \cancel{10^{23}} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 6.022 \times 10^{22}+ 2\times 6.022 \times 10^{22} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 6.022 \times 10^{22}+ 12.044 \times 10^{22} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf (6.022+12.044) \times 10^{22}\\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 18.066 \times 10^{22}\\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 1.8066 \times 10^{23}\\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 1.8 \times 10^{23}\quad(1.8066 \approx 1.8)\\\end{gathered} [/tex]
The total number of ions in 7.41 g of calcium hydroxide, [tex]\sf Ca(OH)_2 = 1.8 \times 10^{23}[/tex]
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