Answer
$$1.39 \space M \space ZnCl_2$$
Work Step by Step
1. Calculate the amount of moles of $ Zn $:
$ Zn $ : 65.41 g/mol
$$ 25.0 \space g \space Zn \times \frac{1 \space mol \space Zn }{ 65.41 \space g \space Zn } = 0.382\underline{2} \space mol \space Zn $$
2. Find the amount of moles of $ ZnCl_2 $ necessary:
$$ 0.382\underline{2} \times \frac{ 1 \space mol \space ZnCl_2 }{ 1 \space mol \space Zn }= 0.382\underline{2} \space mol \space ZnCl_2 $$
3. Find the molarity of $ ZnCl_2 $:
$$Molarity = \frac{ 0.382\underline{2} \space mol \space ZnCl_2 }{ 275 \space mL \space ZnCl_2 } \times \frac{1000 \space mL}{1 \space L} = 1.39 \space M \space ZnCl_2 $$
