Answer: a) 0.06 M
b) 3.13 grams.
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]
Moles of [tex]Cl^-=Molarity\times {\text {Volume in L}}=0.240\times 0.2L=0.048moles[/tex]
The balanced reaction for dissociation will be:
[tex]SnCl_4\rightarrow Sn^{4+}+4Cl^{-}[/tex]
Thus for 4 moles of [tex]Cl^-[/tex], there is 1 mole of [tex]SnCl_4[/tex]
Thus moles of [tex]SnCl_4=\frac{1}{4}\times 0.048=0.012[/tex]
Molarity of [tex]SnCl_4=\frac{moles}{\text {Volume in L}}=\frac{0.012}{0.2L}=0.06M[/tex]
Thus the molarity of the [tex]SnCl_4[/tex] solution = 0.06 M
b) Mass of [tex]SnCl_4[/tex]= [tex]moles\times {\text {Molar mass}}=0.012mol\times 260.522g/mol=3.13g[/tex]
Thus mass of [tex]SnCl_4[/tex] Wenzhou use is 3.13 grams.
Answer:
a) [tex]M_{SnCl_4}=0.06M[/tex]
b) [tex]m_{SnCl_4}=3.126gSnCl_4[/tex]
Explanation:
Hello,
a) In this case, as ideally we have 0.240 moles of chloride ions per liter of solution, one computes the actual molarity of SnCl₄ as shown below via dimensional analysis:
[tex]M_{SnCl_4}=0.240\frac{molCl^-}{1Lsln} *\frac{1molSnCl_4}{4molCl^-}=0.06\frac{molSnCl_4}{L} =0.06M[/tex]
b) Now, since Wenzhou prepared 200 mL, the used mass is computed as follows, in which the molar mass of SnCl₄ is considered:
[tex]m_{SnCl_4}=0.06\frac{molSnCl_4}{Lsln}*0.200Lsln*\frac{260.5gSnCl_4}{1molSnCl_4} \\m_{SnCl_4}=3.126gSnCl_4[/tex]
Best regards.
