what is the molarity of that forms when 25.0 g of zinc completely reacts with according to the following reaction? assume a final volume of 275 ml

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.

i think its 5049.

i hope this !