Question

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

Answer 1

`[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"`........

Answer 2

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 1/2`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.

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