Question

4. Calculate the molality of a 20% w/w aqueous solution of sulphuric acid, H2SO4? (Atomic masses are: S= 32 amu, H = 1 amu, O = amu)

Answer 1

Molecular weight of sulphuric acid `=98 "g/mol"`

There's `20 g` acid in `100 g` of water.

Thus actual mass of water present is `80 g`.

Molality means number of solute particles present per `kg` of solvent.

Thus,

`80g` of solvent(water) has `20 g` acid

`=> 1g` of solvent has `(20)/(80)g` acid.

`=> 1000g(=1kg)` of solvent has `(20)/(80)×1000= 250g` of acid.

Now,

`98g` of acid means `1` mole of acid.

`=>1g` of acid means `1/(98)` moles of acid.

`=>250g` of acid means `1/(98)×250= 2.55` moles of acid.

Thus, molality of the solution is 2.55 molal because 2.55 moles of acid are present in 1000g of water

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Or going by the conventional method, and applying the formula,

Molality`=("given mass")/("molar mass")×1000/("mass of solvent in g")`

`=(20)/(98)×(1000)/(80) = 2.55 " mol/kg"`

Answer 2

You need to quote the density of the acid solution....

Explanation:

We wants `"molality"="moles of sulfuric acid solute"/"kilograms of water solvent"`. Now this site quotes `rho_"20% sulfuric acid"=1.21*g*mL^-1`....and so we work from a `1*mL` volume of this solution....the which of course has a mass of `1.21*g`...

`"molality"=((20%xx1.21*g*mL^-1xx1*mL)/(98.08*g*mol^-1))/(80%xx1.21*gxx10^-3*kg*g^-1)=(2.47xx10^-3*mol)/(9.68xx10^-4*kg)=2.55*mol*kg^-1`