Answer: The molarity of iron(II) sulfate solution is 0.724 M
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Given mass of iron(II) sulfate = 2.75 g
Molar mass of iron(II) sulfate = 152 g/mol
Volume of solution = 150.0 mL
Putting values in above equation, we get:
[tex]\text{Molarity of iron(II) sulfate}=\frac{2.75\times 1000}{152\times 25.0}\\\\\text{Molarity of iron(II) sulfate}=0.724M[/tex]
Hence, the molarity of iron(II) sulfate solution is 0.724 M
