Answer: The molarity of iron(II) sulfate solution is 0.724 M
Step-by-step explanation:
To calculate the molarity of solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}](https://img.qammunity.org/2021/formulas/chemistry/college/d4xdoph0eex2l3cldfyerzzxfd1m1qvaw8.png)
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:
![\text{Molarity of iron(II) sulfate}=(2.75* 1000)/(152* 25.0)\\\\\text{Molarity of iron(II) sulfate}=0.724M](https://img.qammunity.org/2021/formulas/chemistry/college/iyje5oftm5n9uava24tnavafzljrxady56.png)
Hence, the molarity of iron(II) sulfate solution is 0.724 M