Question

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

Answer 1

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]