Question

Calculate molarity, molality and normality of 98% aqueous solution of H2SO4.
(mass/volume) and (density = 1.19g/cm-3)

Answer 1

There is a straight formula for converting w/w% into molarity.


Molarity={(w/w%)×10×(density in g/ml)}÷(Molar mass of solute in g)


Substituting values:

M=(98×10×1.19)/(98)= 11.9 M

Now: Normality=Molarity×n-factor

n -factor is 2 because H2SO4 is a dibasic acid)


Substituting values:N=11.9 × 2= 23.8

Molality=(1000×Molarity)/{(1000×density in g/ml)-(Molar mass of solute in g×Molarity)}

Substituting values,

m=(1000×11.9 )/{(1000×1.19) -(98×11.9 )}

=11900/(1190 – 1166.2)

=11900/23.8=500 n

Answer 2

Answer:

Explanation:

There is a straight formula for converting w/w% into molarity.

Molarity={(w/w%)×10×(density in g/ml)}÷(Molar mass of solute in g)

Substituting values:

M=(98×10×1.19)/(98)= 11.9 M

Now: Normality=Molarity×n-factor

n -factor is 2 because H2SO4 is a dibasic acid)

Substituting values:N=11.9 × 2= 23.8

Molality=(1000×Molarity)/{(1000×density in g/ml)-(Molar mass of solute in g×Molarity)}

Substituting values,

m=(1000×11.9 )/{(1000×1.19) -(98×11.9 )}

=11900/(1190 – 1166.2)

=11900/23.8=500 n