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Question: What is the molality of a 1M NaOH solution? [density of NaOH solution=\(1.04gm{{l}^{-1}}\) ]...

What is the molality of a 1M NaOH solution?
[density of NaOH solution=1.04gml11.04gm{{l}^{-1}} ]

Explanation

Solution

First of all by using the density formula, we will find the mass of the solvent in kilograms and then by applying the molality formula as;-Molality=No. of moles of the solutemass of the solvent in kilogramsMolality=\dfrac{No.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{mass\text{ }of\text{ }the\text{ }solvent\ in\text{ kilo}grams}, we can easily calculate the molality of the NaOH solution. now solve it.

Complete answer:
First of all, let’s discuss what is molality. Molality may be defined as the number of moles of the solute to the mass of solvent in kilograms. It is denoted by m.
Molality of the solution is always independent of the temperature and does not change with the change in the temperature because the mass of the solvent is independent of temperature and hence, the molality.
Mathematically;
Molality=No. of moles of the solutemass of the solvent in kilogramsMolality=\dfrac{No.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{mass\text{ }of\text{ }the\text{ }solvent\ in\text{ kilo}grams}
Now considering the statement as;
1m NaOH means 1mole of NaOH is present in 1L of the solution.
First, we have to find the mass of solvent i.e. the water by using the density formula.
we know that;
volume of solution of NaOH= 1 L =1000 ml (given)
density of NaOH solution=1.04gml11.04gm{{l}^{-1}}(given)
now applying the density formula as;
density=mass of the solutionvolume of the solution 1.04=mass of the solution1000 mass of the solution=1.04×1000  =1040g \begin{aligned} & density=\dfrac{mass\text{ }of\text{ }the\text{ solution}}{\text{volume }of\text{ }the\text{ solution}} \\\ & 1.04=\dfrac{mass\text{ }of\text{ }the\text{ solution}}{1000} \\\ & mass\text{ }of\text{ }the\text{ solution}=1.04\times 1000 \\\ & \text{ =1040g} \\\ \end{aligned}
Thus, the mass of water= 104040g=1000g=1.0kg1040-40g=1000g=1.0kg
Moles of the solute=1
now applying the molarity formula as;
Molality=No. of moles of the solutemass of the solvent in kg  =11.0  =1 mole kg1 \begin{aligned} & Molality=\dfrac{No.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{mass\text{ }of\text{ }the\text{ }solvent\ in\text{ kg}} \\\ & \text{ =}\dfrac{1}{1.0} \\\ & \text{ =1 mole k}{{\text{g}}^{-1}} \\\ \end{aligned}
Thus, the molality of a 1M NaOH solution having the density of NaOH solution as 1.04gml11.04gm{{l}^{-1}} is 1mole kg11mole\text{ }k{{g}^{-1}}.

Note:
Don’t get confused in the terms molarity and the molality. By the term, molarity, we mean the number of moles of the solute to the total volume of the solution in liters. It is denoted by M. whereas on the other hand, by the term molality , we mean the number of moles of the solute to the total mass of solvent in kilograms. It is denoted by m.