A bottle 1litre capacity is labelled as 1 molar Al2(SO4)3 if the bottle is half filled and density of solution is 1.342g/mL molality of Al3 in this solution will be?

Question

Answer

Calculation of Molality of Al3 in the Solution

Given:

Capacity of bottle = 1 litre

Labelled concentration = 1 Molar

Density of solution = 1.342 g/mL

Step 1: Calculation of Number of Moles of Al2(SO4)3 in the Solution

Number of moles = Molarity x Volume

Volume of solution = 1/2 x 1 litre = 0.5 litre

Number of moles of Al2(SO4)3 = 1 Molar x 0.5 litre = 0.5 moles

Step 2: Calculation of Mass of Al2(SO4)3 in the Solution

Mass = Number of moles x Molar mass

Molar mass of Al2(SO4)3 = (2 x Atomic mass of Al) + (3 x Atomic mass of S) + (12 x Atomic mass of O)

Molar mass of Al2(SO4)3 = (2 x 27) + (3 x 32) + (12 x 16) = 342 g/mol

Mass of Al2(SO4)3 in the solution = 0.5 moles x 342 g/mol = 171 g

Step 3: Calculation of Mass of Solvent (Water) in the Solution

Mass of solution = Volume x Density

Mass of solution = 0.5 litre x 1.342 g/mL = 671 g

Mass of solute (Al2(SO4)3) = 171 g

Mass of solvent (Water) = Mass of solution - Mass of solute

Mass of solvent (Water) = 671 g - 171 g = 500 g

Step 4: Calculation of Molality of Al3 in the Solution

Molality = Number of moles of solute / Mass of solvent (in kg)

Number of moles of Al3 = 2 x Number of moles of Al2(SO4)3 (Since 2 moles of Al3 are present in 1 mole of Al2(SO4)3)

Number of moles of Al3 = 2 x 0.5 moles = 1 mole

Mass of solvent (Water) in kg = 500 g / 1000 = 0.5 kg

Molality of Al3 = 1 mole / 0.5 kg = 2 mol/kg

Final Answer: The molality of Al3 in the solution is 2 mol/kg.

Answer

Let the moles of solution be - 2, and volume be - 2L
so that,
2moles/2L = 1 M solution

now, half filled means the moles and volume are reduced to half of there initial values i.e. 1 mole and 1L

density = Mass(g)/Vol (mL)
1.342 = mass(g) / 1×1000(ml)
mass = 1342g

molality = moles/mass of solvent

mass of solvent = 1342 - 342 (mass of 1 mole of Al2(SO4)3 )

molality = 1/1342-342/1000
{dividing by 1000 to make it kg}

molality of Al2(SO4)3 = 1
for Al³ = atomicity × Molality
= 2 × 1
hence, the answer is 2

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