Respuesta :
Answer:
[tex] 1.668\times 10^{12} [/tex] grams of mercury(I) chloride are needed to saturate [tex]4900 km^3[/tex] of water .
Explanation:
[tex]Hg_2Cl_2\rightleftharpoons Hg_2^{2+}+2Cl^-[/tex]
          S    2S
Solubility product of mercury (I) chloride= [tex]K_{sp}=1.5\times 10^{-18}[/tex]
The expression of solubility product is given as:
[tex]K_{sp}=[Hg_2^{2+}][Cl^-]^2[/tex]
[tex]K_{sp}=[S][2S]^2=4S^2[/tex]
[tex]1.5\times 10^{-18}=4S^3[/tex]
[tex]S= 7.211\times 10^{-7} M[/tex]
Volume of water  V = [tex]4900 km^3[/tex]
[tex]km^3=10^{12} L[/tex]
[tex]V=4900\times 10^{12} L[/tex]
Moles of mercury(I) chloride in solution= n
[tex]n=S\times V=7.211\times 10^{-7} M\times 4900\times 10^{12} L[/tex]
Mass of n moles of mercury(I) chloride: m
[tex]=7.211\times 10^{-7} M\times 4900\times 10^{12} L\times 472 g/mol[/tex]
[tex]m= 1.668\times 10^{12} g[/tex]
[tex] 1.668\times 10^{12} [/tex] grams of mercury(I) chloride are needed to saturate [tex]4900 km^3[/tex] of water .
