Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
