What are #"normality"# and #"molality"# for a #"36% (m/m)"# solution of hydrochloric acid, that has #rho_"solution"=1.180*g*mL^-1#?

For #"molarity"#, we want the quotient....

#"Moles of solute"/"Volume of solution"#...and so WE WORK FROM a VOLUME of #1*mL#.

#[HCl]=((1*mLxx1.180*g*mL^-1xx36%)/(36.46*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=11.7*mol*L^-1#.

And #"normality"# is the same as this value; because #HCl# is monoprotic...

For #"molality"# we want the quotient....#"Moles of solute"/"Kilograms of solvent"#

...and again we work from a #1*mL# volume, but we have to pfaff about with units....in particular we have to find the MASS of the water....hence the expression in the denominator; the solute mass (the which I calculated seperately from the percentage) is SUBTRACTED from the mass of solution...to give the mass of the solvent.

#HCl_"molality"=((1*mLxx1.180*g*mL^-1xx36%)/(36.46*g*mol^-1))/((1.180*g-0.425*g)xx10^-3*kg*g^-1)=15.4*mol*kg^-1#

At lower concentrations, #"molality"-="molarity"#........

Normality is the mass of substance delivering 1 mole of cationic charge in a metathesis reaction. For #H_2SO_4#. From this,
=> 1 mole ##H_2SO_4# => 2 "mole" H^+, or 1/2#mole #H_2SO_4# => 1 mole of #H^+# ions. Therefore, 1 equivalent weight of #H_2SO_4# = #98/2 = 49 g/eqv. Wt.#

Normality = #("No." of Eqv Wts)/(Liter of Solution)#

Using the 11.7M #H_2SO_4#,

=> Normality = #((11.7"mole")/L)(98g)/("mole")((1"eqv.""wt.")/(49g))# = 8.6Normal Soluton.

Short cut:
=> In general, Normality of a given solution = #"Molarity"#/#"number of cations"#.