Question

the density of 3 molal solution of NAOH is 1.11 g/mL .calculate the molarity of the solution

Answer 1

A 3 molal solution is prepared by dissolving 3 moles (40 x 3 = 120g) of NaOH in 1000g 0f water.

Total mass of such solution containing 3 moles of the solute = 1000 +120 = 1120g

Density of this solution = 1.110g/ml

Density = mass/volume

Total volume of this solution = 1120/1.110 = 1009mL = 1.009L

Thus, 3 moles of NaOH are present in 1.009L of the solution.

So, no of moles present in 1L of the solution = 3 x 1/1.009 = 2.973

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Answer 2

Find Molarity , given Molality.

Molality = b = 3

mass density of Sodium Hydroxide solution = ρ = 1.110 gm/ml 

     ρ = 1.110 kg/Litre

M = Molar mass of Sodium Hydroxide NaOH = 40 = (23+16+1)

     = 40 gm/mole = 0.040 kg/mole

Molarity = c = ??

Formula for calculating Molarity given Molality:

          c = ρ * b / [1 + b M ]

             = 1.110 kg/L * 3 mole/kg  / [ 1 + 3 mole/kg * 0.040 kg/mole ]

             = 3.330 /(1.12) 

             = 2.97 Molar

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