Question

Wenzhou prepares 200 mL of a solution of SnCl4in which the concentration ofchloride ions is 0.240M.a) What is the molarity of the SnCl4solution (i.e. what should the bottle be labeled)?b) What mass of SnCl4did Wenzhou use?

Answer 1

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.

$Molarity=\frac{moles}{\text {Volume in L}}$

Moles of  $Cl^-=Molarity\times {\text {Volume in L}}=0.240\times 0.2L=0.048moles$

The balanced reaction for dissociation will be:

$SnCl_4\rightarrow Sn^{4+}+4Cl^{-}$

Thus for 4 moles of $Cl^-$, there is 1 mole of $SnCl_4$

Thus moles of $SnCl_4=\frac{1}{4}\times 0.048=0.012$

Molarity of $SnCl_4=\frac{moles}{\text {Volume in L}}=\frac{0.012}{0.2L}=0.06M$

Thus the molarity of the  $SnCl_4$ solution = 0.06 M

b) Mass of $SnCl_4$= $moles\times {\text {Molar mass}}=0.012mol\times 260.522g/mol=3.13g$

Thus mass of  $SnCl_4$ Wenzhou use is 3.13 grams.

Answer 2

Answer:

a) $M_{SnCl_4}=0.06M$

b) $m_{SnCl_4}=3.126gSnCl_4$

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:

$M_{SnCl_4}=0.240\frac{molCl^-}{1Lsln} *\frac{1molSnCl_4}{4molCl^-}=0.06\frac{molSnCl_4}{L} =0.06M$

b) Now, since Wenzhou prepared 200 mL, the used mass is computed as follows, in which the molar mass of SnCl₄ is considered:

$m_{SnCl_4}=0.06\frac{molSnCl_4}{Lsln}*0.200Lsln*\frac{260.5gSnCl_4}{1molSnCl_4} \\m_{SnCl_4}=3.126gSnCl_4$

Best regards.