Answer
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Hint: Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is mol per kg. Molality is an intensive property of solutions, and it is calculated as the moles of a solute divided by the kilograms of the solvent.
Complete step by step answer:
It is given in the question that the density of the solution is $1.25\,gm{{L}^{-1}}$ and the molarity of the solution is $3\,mol\,{{L}^{-1}}$. Using these two informations we will calculate the Molality of the solution.
Now, we will find the mass of the NaCl present in the solution.
Molar Mass of \[NaCl=58.5g\,mo{{l}^{-1}}\]
We know that,
No. of moles = $\dfrac{Given\,mass}{Molar\,mass}$
$Given\,mass=Number\,of\,Moles\times Molar\,mass$
Mass of NaCl in 1-Liter solution = $3\times 58.5=175.5g$
Now, we will find the solution. To calculate the mass of water we will multiply the density of the solution with the volume of the solution that is 1000mL. This will give us the mass of 1 litre solution.
\[Density=\dfrac{Mass}{Volume}\]
Therefore, $Mass\,=\,Density\times \,Volume$
Mass of 1 liter solution = $1000\times 1.25=1250g$
Now, we know that the mass of the solution is the sum of the mass of the solvent and the mass of solute. In this case the solvent is water and the solute is NaCl.
So, the mass of the water will be equal to the difference of the mass of the solution and the solute.
Therefore,
Mass of water in solution = Mass of solution – Mass of solute = 1250-175.5 = 1074.5 g
Mass of water in solution = 1.074 kg
Now, we will finally calculate the molality of the solution. To calculate the molality of the solution we should know the moles of the solute and the mass of the solvent in Kg.
\[Molality\,of\,solution=\dfrac{Number\,of\,Moles\,of\,solute}{Mass\,of\,solvent\,(in\,kg)}\]
\[Molality\,of\,solution=\dfrac{3}{1.074}\]
\[Molality=2.79m\]
The molality of the solution is 2.79 m.
Note:
It is easy to calculate molality if we know the mass of solute and solvent in a solution. Molality is an intensive property and is therefore independent of the amount being measured. This is true for all homogeneous solution concentrations, regardless of if we examine a 1.0 L or 10.0 L sample of the same solution. The concentration, or molality, remains constant.
Complete step by step answer:
It is given in the question that the density of the solution is $1.25\,gm{{L}^{-1}}$ and the molarity of the solution is $3\,mol\,{{L}^{-1}}$. Using these two informations we will calculate the Molality of the solution.
Now, we will find the mass of the NaCl present in the solution.
Molar Mass of \[NaCl=58.5g\,mo{{l}^{-1}}\]
We know that,
No. of moles = $\dfrac{Given\,mass}{Molar\,mass}$
$Given\,mass=Number\,of\,Moles\times Molar\,mass$
Mass of NaCl in 1-Liter solution = $3\times 58.5=175.5g$
Now, we will find the solution. To calculate the mass of water we will multiply the density of the solution with the volume of the solution that is 1000mL. This will give us the mass of 1 litre solution.
\[Density=\dfrac{Mass}{Volume}\]
Therefore, $Mass\,=\,Density\times \,Volume$
Mass of 1 liter solution = $1000\times 1.25=1250g$
Now, we know that the mass of the solution is the sum of the mass of the solvent and the mass of solute. In this case the solvent is water and the solute is NaCl.
So, the mass of the water will be equal to the difference of the mass of the solution and the solute.
Therefore,
Mass of water in solution = Mass of solution – Mass of solute = 1250-175.5 = 1074.5 g
Mass of water in solution = 1.074 kg
Now, we will finally calculate the molality of the solution. To calculate the molality of the solution we should know the moles of the solute and the mass of the solvent in Kg.
\[Molality\,of\,solution=\dfrac{Number\,of\,Moles\,of\,solute}{Mass\,of\,solvent\,(in\,kg)}\]
\[Molality\,of\,solution=\dfrac{3}{1.074}\]
\[Molality=2.79m\]
The molality of the solution is 2.79 m.
Note:
It is easy to calculate molality if we know the mass of solute and solvent in a solution. Molality is an intensive property and is therefore independent of the amount being measured. This is true for all homogeneous solution concentrations, regardless of if we examine a 1.0 L or 10.0 L sample of the same solution. The concentration, or molality, remains constant.
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