A sample of 12 M concentrated hydrochloric acid has a density 1.2 ` g L ^(-1) ` Calculated the molality.

Explanation

Given: Molarity `= 12 MHCl`
Density of the solution `=1.2 g L ^(-1)`
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Molality `= ("No. of moles of solute")/("Mass of solvent (in kg)")`
Caculate mass of water (solvent)
Mass of 1 litre HCl solution = density `xx` volume
`=1.2 gm L ^(-1) xx 100 mL = 1200g`
Mass of HCl = No of moles of HCl` xx` molar mass of HCl
Mass of water = mass of HCl solution - mass of HCl
Mass of water `=1200 - 438 =762 g`
Molality `= (12)/(0.762) =15.75m`