The density of 3 M solution of NaCl is 1.25 g mL-1. Calculate molality

1.	The density of 3 M solution of NaCl is 1.25 g mL-1. Calculate molality of the solution.
Answer» Given : Molarity of the solution = 3 M, Density of the solution = 1.25 g mL To find : Molality of the solution Formula : Molality = \(\frac{Number\,of\,moles\,of\,solute}{Mass\,of\,solvent\,in\,kilograms}\) Calculation : Molarity = 3 mol L^-1 ∴ Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g Mass of 1 L solution = 1000 × 1.25 = 1250 g (∵ Density = 1.25 g mL^-1) Mass of water in solution = 1250 – 175.5 = 1074.5 g = 1.0745 kg Molality = \(\frac{Number\,of\,moles\,of\,solute}{Mass\,of\,solvent\,in\,kilograms}\) = \(\frac{3\,mol}{1.0745\,kg}\) = 2.790 m (by using log table) ∴ Molality of the NaCl solution = 2.790 m [Calculation using log table : \(\frac{3}{1.0745}\) = Antilog₁₀[log₁₀(3) – log₁₀(1.0745)] = Antilog₁₀[0.4771 – 0.0315] = Antilog₁₀[0.4456] = 2.790]

The density of 3 M solution of NaCl is 1.25 g mL-1. Calculate molality of the solution.

Answer»

Given :

Molarity of the solution = 3 M,

Density of the solution = 1.25 g mL

To find :

Molality of the solution

Formula :

Molality = \(\frac{Number\,of\,moles\,of\,solute}{Mass\,of\,solvent\,in\,kilograms}\)

Calculation :

Molarity = 3 mol L^-1

∴ Mass of NaCl in 1 L solution

= 3 × 58.5

= 175.5 g

Mass of 1 L solution = 1000 × 1.25

= 1250 g

(∵ Density = 1.25 g mL^-1)

Mass of water in solution = 1250 – 175.5

= 1074.5 g

= 1.0745 kg

Molality = \(\frac{Number\,of\,moles\,of\,solute}{Mass\,of\,solvent\,in\,kilograms}\)

= \(\frac{3\,mol}{1.0745\,kg}\)

= 2.790 m (by using log table)

∴ Molality of the NaCl solution = 2.790 m

[Calculation using log table : \(\frac{3}{1.0745}\)

= Antilog₁₀[log₁₀(3) – log₁₀(1.0745)]

= Antilog₁₀[0.4771 – 0.0315]

= Antilog₁₀[0.4456] = 2.790]