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

2 Answers
Nov 17, 2017

[HCl]=11.7*mol*L^-1....

Explanation:

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"........

Nov 17, 2017

Normality of the H_2SO_4(aq) solution = 8.6 Normal (=> 8.6N)

Explanation:

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 1/2mole 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".