Density of 2 m aqueous solution of naoh is 1.28 g/cm3. The molality of the solution is [given that molecular mass of naoh = 40 g molâ1]
Answers
Explanation:
The given data is as follows.
Molarity = 2 M, Density = 1.28 = 0.00128 g/L (as 1 = 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 =
1280 =
mass = 1280 g
As it is known that, number of moles =
Therefore, mass of solute (NaOH) is as follows.
number of moles =
2 =
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, = = 1.2 kg
Now, calculate the molality as follows.
Molality =
=
= 1.6 mol/kg
Thus, we can conclude that molality of the given solution is 1.6 mol/kg.
Given that, the density of 2 M aqueous solution of NaOH is 1.28 g/cm³.
{ Density (d) is 1.28 g/cm³ and Molarity (M) is 2M }
We have to find molality (m) of the solution.
Now,
The relation between Molality, Molarity and Density is:
m = 1000M/(1000d - 2m.m.)
Here, m = molality, M = Molarity, d = density and m.m. = molar mass
Also given that, Molar Mass (m.m.) of NaOH = 40 g/mol
Substitute the known values in the above formula,
→ m = [1000(2)]/[1000(1.28) - 2(40)]
→ m = 2000/(1280.00 - 80)
→ m = 2000/(1280 - 80)
→ m = 2000/1200
→ m = 20/12
→ m = 5/3
→ m = 1.67
Therefore, the molality of the solution is 1.67 Molal.