250 ml of 0.1 M Fe(NH_{4})_{2}(SO_{4})_{2}×6H_{2}O:

molar mass Fe(NH_{4})_{2}(SO_{4})_{2}×6H_{2}O = 392.12 g/mol

250 ml x 1 L/1000 ml = 0.25 L

0.1 M = 0.1 mol/L

0.1 mol/L x 0.25 L = 0.025 mols needed

0.025 mols x 392.12 g / mol = 9.803 g Fe(NH_{4})_{2}(SO_{4})_{2}×6H_{2}O needed = 9.8 g (2 sig. figs.)

Weigh out 9.8 g Fe(NH_{4})_{2}(SO_{4})_{2}×6H_{2}O and dissolve in a final volume of 250 ml of 5% v/v H_{2}SO_{4}

iron(II) ammonium sulphate hexahydrate = (NH_{4})_{2}Fe(SO_{4})_{2} • 6H_{2}O molar mass = 392.14 g/mol

250.0 ml x 1 L / 1000 ml = 0.2500 L

0.01 M = 0.01 mols / L

0.01 mol/L x 0.2500 L = 0.002500 mols (NH_{4})_{2}Fe(SO_{4})_{2} • 6H_{2}O

0.002500 mols x 392.14 g/mol = 0.9804 g (NH_{4})_{2}Fe(SO_{4})_{2} • 6H_{2}O needed

The question is a little unusual as it first asks to prepare the solution accurately but then is uses ca. 0.01 M which means it approximately 0.01 M, so what is the level of accuracy? Furthermore, it asks to prepare 250.0 mls but then says to use 200 ml od 5% v/v sulfuric acid. Maybe this was a typo and it means to use 250 mls of sulfuric acid as in the previous question. That being the case it would be...

weigh 0.98 g of (NH_{4})_{2}Fe(SO_{4})_{2} • 6H_{2}O and dissolve in a final volume of 250 mls 5% v/v H_{2}SO_{4}