**Density of a 2M Aqueous Solution of NaOH**

**Introduction:**
Density is a physical property that describes the mass of a substance per unit volume. In the case of a solution, the density is influenced by the concentration of the solute and the solvent. In this explanation, we will discuss the density of a 2M aqueous solution of sodium hydroxide (NaOH).

**Definition of Molarity:**
Molarity (M) is a unit of concentration that represents the number of moles of solute per liter of solution. In the case of a 2M NaOH solution, it means that there are 2 moles of NaOH dissolved in 1 liter of water.

**Determining the Density:**
To determine the density of a 2M NaOH solution, we need to consider the masses and volumes of both the solute (NaOH) and the solvent (water). The density formula is given by:

Density = Mass / Volume

**Mass Calculation:**
The mass of the solution is the sum of the masses of the solute (NaOH) and the solvent (water). However, since the density is given in grams per cubic centimeter (g/cm^3), it is convenient to convert the volume from liters to cubic centimeters.

**Volume Calculation:**
The volume of the solution can be calculated by dividing the mass of the solution by its density. In this case, we are given that the density of the 2M NaOH solution is 1.28 g/cm^3.

**Calculation:**
Let's assume we have 1 liter (1000 cm^3) of the 2M NaOH solution.

Given:
Molarity (M) = 2M
Density = 1.28 g/cm^3
Volume = 1000 cm^3

**Mass Calculation:**
Mass of NaOH = Moles of NaOH * Molar Mass of NaOH
Molar Mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol
Moles of NaOH = Molarity * Volume (in liters)
Moles of NaOH = 2M * 1L = 2 moles
Mass of NaOH = 2 moles * 39.99 g/mol = 79.98 g

**Volume Calculation:**
Density = Mass / Volume
1.28 g/cm^3 = 79.98 g / Volume
Volume = 79.98 g / 1.28 g/cm^3 = 62.48 cm^3

Therefore, the mass of the 2M NaOH solution is 79.98 grams, and the volume is 62.48 cm^3.

**Conclusion:**
The density of a 2M aqueous solution of NaOH is 1.28 g/cm^3. This value is obtained by calculating the mass and volume of the solution. The mass is determined by considering the moles of NaOH and its molar mass, while the volume is calculated using the given density.

The given data is as follows.

Molarity = 2 M, Density = 1.28 g/cm^{3}g/cm
3
= 0.00128 g/L (as 1 cm^{3}cm
3
= 0.001 L)

Molar mass = 40 g/mol

Therefore, it means that there are 2 moles present in 1 liter. Hence, calculate the mass of solution as follows.

Density = \frac{mass}{volume}
volume
mass
​


1280 g/Lg/L = \frac{mass}{1 L}
1L
mass
​


mass = 1280 g

As it is known that, number of moles = \frac{mass}{Molar mass}
Molarmass
mass
​


Therefore, mass of solute (NaOH) is as follows.

number of moles = \frac{mass}{Molar mass}
Molarmass
mass
​


2 = \frac{mass}{40 g/mol}
40g/mol
mass
​


mass = 80 g/mol

Now, mass of solute + mass of solvent = mass of solution.

80 g + mass of solvent = 1280 g

mass of solvent = 1280 g - 80 g

= 1200 g

or, = \frac{1200}{1000}
1000
1200
​
= 1.2 kg

Now, calculate the molality as follows.

Molality = \frac{\text{no. of moles of solute}}{\text{mass of solution in kg}}
mass of solution in kg
no. of moles of solute
​


= \frac{2}{1.2}
1.2
2
​


= 1.6 mol/kg

Thus, we can conclude that molality of the given solution is 1.6 mol/kg.