Question

Calculate : Molarity , Molality & normality
of 32 % by maas H2SO4
solution having
density 1.086 g/cm3​

Answer 1

Let the mass of solution be 100gms

then the wieght if H2SO4 will be

32gm

Moles of h2so4 = mass÷molecular mass

$Moles_{H_{2}SO_{4}}=\frac{mass\: of\: solute}{Molecular\:mass}$

$Moles=\frac{32}{98}=\bold{0.33Moles}$

Now We know that Weight of solution is 100gms

also given

density=1.086gms

$Density(\rho)=\frac{mass}{volume}$

$\rho =\frac{100}{Volume}$

$Volume=\frac{100}{\rho}$

$Volume=\frac{100}{1.086}=\pink{92.1 ml}$

$Volume=92.1 ml or \bold{92.1×10^{-3}L}$

We know that nfactor of H2SO4 is 2 as Net positive or negative charge over it is 2

Also that weight of solution is 100gms and

weight of solute is 32 gms

Then weight of solvent is 68gms or 0.068kg

Now using the formulas

$Molarity=\frac{Moles_{solute}}{Volume_{solution}</p><p> \:in \:Litre}$

$Molality=\frac{Moles_{solute}}{Mass_{Solvent}\:in \: kg}$

$Normality=Molarity×nfactor$

$Molarity=\frac{0.33}{92.1×10^{-3}}=\bold{3.6M}</p><p>$$Molality=\frac{0.33}{0.068}=\bold{4.9m}$

$Normality=M×nf=3.6×2=\bold{7.2N}$