Question

A solution is prepared by adding .8 g of iron sulfate to 450mL of water

Answer 1

Molar mass FeSO₄ =151.908 g/mol

Volume = 450 mL / 1000 = 0.45 L

1 mol ----------- 151.908 g
n mol ----------- 0.8 g

n = 0.8 * 1 / 151.908

n = 0.00526 moles 

M = n / V

M = 0.00526 / 0.45

M = 0.01168 mol/L⁻¹

hope this helps!