Answer:
Molarity of \mathrm{ZnCl}_{2}ZnCl
2
is 1.364 M
Explanation:
Molarity of product Zinc Chloride which is formed when 25.0 g of Zinc completely reacts with copper(II) chloride: \mathbf{Z n}(\mathbf{s})+\mathbf{C u C l}_{2}(\mathbf{a q}) \rightarrow \mathbf{Z n C l}_{2}(\mathbf{a q})+\mathbf{C u}(\mathbf{s})Zn(s)+CuCl
2
(aq)→ZnCl
2
(aq)+Cu(s)
Firstly no of moles of Zn can be obtained from :
\text { Moles of } \mathbf{Z n}=\frac{\text { Mass of Zn }}{\text { Molar Mass of Zn }} Moles of Zn=
Molar Mass of Zn
Mass of Zn
, molar mass of Zn = 65.38 g/mol
hence moles of Zn = 25.0 g / 65.38 g/mol = 0.382 mol.
Now some conversion are needed to be done like:
1 m L = 0.001 L
Volume of solution = 285 mL
Hence 285 mL = 285 × 0.001 L = 0.28 L
Finally Molarity of \mathrm{ZnCl}_{2}ZnCl
2
can be calculated by:
Molarity of \mathrm{ZnCl}_{2}ZnCl
2
= \frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}
Volume of solution in litres
number of moles of solute
Molarity = \frac{\text { 0.382 mol}}{\text { 0.28 L }}
0.28 L
0.382 mol
= 1.364 M.
