The density of 3M solution of NaCL is 1.25 g`mL^(-1)`. Calculate molality of the solution.

Explanation

`M=3"mol"L^(-1)` Mass of NaCl in 1 L solution = 3xx58.5=175.5 g
Mass of
1 L solution = 1000xx1.25=1250g ( Since density = 1.25 `gmL^(-1)` )
Mass of water of solution =1250-175.5
= 1074.5 g=1.0745 kg
Molality = `("No .of moles of solute")/("Mass of solvent. in kg")`
= `(3"mol")/(1.0745 kg)=2.79 .m` Often in a chemistry laboratory. a solution of a desired concentration is prepared by di-luting a solution of known higher concentra-tion. The solution of higher concentration is also known as stock solution. Note that mo-lality of a solution does not change with tem-perature since mass remains unaffeoted with temperature.