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

Explanation

Molality = 12 MHC1
Dentisity of the solution = ` 1.2 gL^(-1)`
In 12 M HCI solution there are 12 moles of HC1 in 1 litre of the solution
` "Molality " =( " No . Of moles of solute ")/( " Mass of the solvent ( in kg ) ") `
Calculated mass of water ( solvent )
Mass of 1 litre HC1 solution = density `xx` volume
` " " = 1.2 gmL^(-1) xx 1000 mL `
` " " =1200g `
Mass of HC1 = no. moles of HC1`xx ` molar mass of HC1
`= 12 " mol " xx36.5 "mol "^(-1) =438 g`
Mass of water = mass of HC1 solution -mass of HCI
Mass of water =` 1200 -438 =762 g`
Molality =` ( 12)/( 0.762) = 15.75m`