Question

What mass of solid Al2(SO4)3 is required to make
41.0 ML of a 0.0800m solution Al2 (S04)3 ?
The answer is​

Answer 1

,

given-

molality = 0.15 m

V(volume) = 100 ml

density = 1.5 gm/ml

as we know

m(molality) = (M x 1000) / (d x 1000 - M x Mw)

here

M = molarity

d = density

Mw = molecular weight

then

0.15 = (M x 1000)/(1.5 x 1000 - M x 342)

225 - 51.3M = 1000M

so

Molarity (M) = 0.214 M

then

we know that

M = (wt/Mw) x (1000/V in ml)

0.214 = (wt/342) x (1000/100)

so wt = 7.319 gm

and

2Al^+3 + 3(SO4)^-2 ----> Al2(SO4)3

and

weight of Al2(SO4)3 is 7.319 gm

then mole of Al2(SO4)3 is 7.32/342 = 0.0214 mole

then by stoichiometry mole of Al^+3 has 2 x 0.0214 = 0.0428 mole

then weight of Al^+3 ion is 0.0428 x 27 = 1.155 gm

and no. of Al^+3 ion is N

and we know one thing

mole = N/NA

N = number of atoms

NA = avogadro number = 6.022 x 10^23

then

0.0428 x 6.022 x 10^23 = N

=> N = 2.57 x 10^22