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Question: Molality of the 18 M \({{H}_{2}}S{{O}_{4}}\) (density = \(1.8\text{ gm }{{\text{L}}^{-1}}\) ) is: - ...

Molality of the 18 M H2SO4{{H}_{2}}S{{O}_{4}} (density = 1.8 gm L11.8\text{ gm }{{\text{L}}^{-1}} ) is: -
(a)- 36 mol/kg
(b)- 200 mol/kg
(c)- 500 mol/kg
(d)- 18 mol/kg

Explanation

Solution

Mass of the solution can be calculated by multiplying the density into the volume of the solution. The molality of the solution is calculated by dividing the moles of the solute to the mass of the solvent in kg.

Complete answer:
Molality: The molality of a solution is defined as the number of moles of the solute dissolved in 1kg (1000 g) of the solvent. It is represented by the symbol, ‘m’.
Molality=Moles of the soluteMass of the solvent in kgMolality=\dfrac{\text{Moles of the solute}}{\text{Mass of the solvent in kg}}
Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.
Molarity=Moles of the soluteVolume of the solutionMolarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}
Moles of the solute is calculated by dividing the given mass of the solute to the molecular mass of the solute.
Moles=mass of the solutemolar mass of the soluteMoles=\dfrac{\text{mass of the solute}}{\text{molar mass of the solute}}
So, the question says 18 M of H2SO4{{H}_{2}}S{{O}_{4}} (sulfuric acid)
It means that 1 L1\text{ }L of the solution contains 18 moles18\text{ }moles of H2SO4{{H}_{2}}S{{O}_{4}} (sulfuric acid)
So, the volume of the solution is 1 L1\text{ }L and the mole of the H2SO4{{H}_{2}}S{{O}_{4}} (sulfuric acid) is1818 .
The density of the solution is given 1.8 gm L11.8\text{ gm }{{\text{L}}^{-1}}, the mass of the solution can be calculated.

The mass of the solution can be calculated by multiplying the density with the volume of the solution.
Mass of the solution = 1 x 1.8 = 1.8 Kg1\text{ x }1.8\text{ }=\text{ }1.8\text{ }Kg
So, the 1 L H2SO4{{H}_{2}}S{{O}_{4}} is equal to 1.8 kg or 1800 g of solution.
So, in the solution the mass of H2SO4{{H}_{2}}S{{O}_{4}} is:
Molarity x molar mass of H2SO4{{H}_{2}}S{{O}_{4}}= 18 x 98 = 1.764 kg.18\text{ x }98\text{ }=\text{ }1.764\text{ }kg.

So, the mass of the solution is 1.8 kg and the mass of H2SO4{{H}_{2}}S{{O}_{4}} is 1.764 kg, so the mass of the solvent is:
=1.81.764=0.036 kg=1.8-1.764=0.036\text{ kg}
So, the molality of the solution is:
Molality=Moles of the soluteMass of the solvent in kg=180.036=500 mMolality=\dfrac{\text{Moles of the solute}}{\text{Mass of the solvent in kg}}=\dfrac{18}{0.036}=500\text{ m}
Hence the molality of the solution is500 m500\text{ }m .
So, the correct answer is “Option C”.

Note: You should always take care that when molality is calculated the mass of the solvent is taken, not the mass of the solution. Don’t get confused that the molarity and molality of the solution are the same.