Question

A sample of commercial concentrated hydrochloric acid is 11.8M HCL and has a density of 1.190g/ml calculate the molality of HCL​

Answer 1

Molarity = 11.8 M

Molarity = (moles of HCl) / (litres of solution)

Molality = (moles of HCl) / (kilograms of solvent)

=> Let number of moles of HCl present be x moles.

So, 11.8 = x/litres of solution

litres of solution = x/11.8 litres.

So, volume in mL = x/11.8 * 1000 mL =>10,000x/118 mL

Density of solution = 1.19 g/ml

So, D = m/v

Mass of solution = D * v => 1.19 * 10,000x/118 => 11,900x/118 g

=> (11,900x/118)/1000 L => 11.9x/118 litres

Moles of HCl = x moles.

Molality = x/( 11.9x/118)

=> 118x/11.9x => 118/11.9 => 9.9156 mol L-1

=> Molality of solution = 9.915 mol L-1 (approx.)