The density of 3 M 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 `= 3 xx 58.5`
`= 175.5 g`
Mass of 1 L solution `= 1000 xx 1.25 = 1250g`
(Since density `= 1.25 g mL^(-1)`)
Mass of water in solution `= 1250-175.5`
`=1074.5g`
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 molality of a solution does not change with temperature since mass remains unaffected with temp.