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

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 = 1250 g`
(since density =1.25 g `mL^(-1)`)
Mass of water in solution `=1250 - 75 .5`
=1074.5 g
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 diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature.