Question

Calculate the molality of 13% its solution by mass of sulphuric acid if density is 1.020

Answer 1

M=~10.20M, m=1.524m

Explanation:

Here,

weight of sulphuric acid=13g(13% out of 100g solution)

Molecular weight=Mw=98

volume=?

density= mass/volume

volume=mass/density

v=13/1.020

v=12.74ml~13

Now,

Molarity(M)= weight of solute×1000

Mw × V(ml)

=13×1000/98×13

=1000/98

=~10.20M

Now,

weight of solute=13g.

We know that,

weight of solution=weight of solute+weight of solvent.

100=13+weight of solvent.

100-13=weight of solvent

87g=weight of solvent.

Now, we know that,

Molality(m) = weight of solute ×1000

Mw × weight of solvent(in g)

= 13×1000/98×87

=13000/98×87

=13000/8526

=1.524m