Determine the molality 0.8 mol/dm3 solution of Fe2 (SO) 4. Density equal to 1 g/cm3.
To determine the molality of the solution, we need to know the amount of solute (in moles) dissolved in a known mass of solvent (in kilograms).
In this case, we are given that the concentration of the solution is 0.8 mol/dm^3, and the density of the solution is 1 g/cm^3.
First, let's convert the concentration from dm^3 to cm^3:
0.8 mol/dm^3 = 0.8 mol/L
Since 1 dm^3 is equal to 1000 cm^3, we can convert the concentration:
0.8 mol/dm^3 = 0.8 mol/L = 0.8 mol/1000 cm^3
Next, we need to calculate the mass of the solute. The formula for iron(II) sulfate is Fe2(SO4)3. Since the molecular weight of Fe2(SO4)3 is 399.88 g/mol, we can calculate the mass of the solute using the following formula:
mass = concentration * volume * molecular weight
Given that the density of the solution is 1 g/cm^3 and the molecular weight of Fe2(SO4)3 is 399.88 g/mol, we can calculate the mass of the solute:
mass = 0.8 mol/1000 cm^3 * 1000 cm^3/L * 399.88 g/mol
mass = 0.8 * 399.88 g
Finally, to determine the molality, we divide the moles of solute by the mass of the solvent in kilograms.
Assuming we have a mass of 1 kg of solvent, we can calculate the molality:
molality = moles of solute / mass of solvent
molality = 0.8 * 399.88 g / 1000 g
Therefore, the molality of the 0.8 mol/dm^3 solution of Fe2(SO4)3 is 0.32 mol/kg.