How many g of Al2*SO4)3 are there in 100 ml of 0.15 m solution of Al2(SO4)3 . The density of the solution is 1.5 g/ml. Also report the no. Of Al+3 ions in this weight .
anonymous22:
molarity =0.15/150-0.15*342/1.5*1000 =1.5M . Hence, 1.5*0.1=moles of solute , therefore wt.=0.15*342=51.3 ,moles of Al+ =0.3 Al+ molecules=0.3Na
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Answered by
7
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
hope it will help you
regards
.
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
hope it will help you
regards
.
Answered by
6
Mass of the solution = volume * density = 100 * 1.5 = 150 gms
Molecular mass of water = 18
Let there be x grams of Aluminum Sulphate Al₂ (SO₄)₃ be present in the solution. Molecular mass of it is 342 = 27*2+32*3+16*12.
moles of Aluminum sulphate = x/342
mass of water in the solution = (150 - x)/1000 kg
Given molality = 0.15 = (x/342) / [(150 - x)/1000]
So 0.15 * (150 - x) = x * 1000/342
x = 7.319 gms
Moles of the sulphate = x/342 = 0.0214
Moles of Al⁺³ ions in the solution = 2 * 0.0214 = 0.0428
Total Number of ions = 0.0428 * 6.023 * 10²³ = 2.578 * 10²²
(multiplication by Avogadro number).
Molecular mass of water = 18
Let there be x grams of Aluminum Sulphate Al₂ (SO₄)₃ be present in the solution. Molecular mass of it is 342 = 27*2+32*3+16*12.
moles of Aluminum sulphate = x/342
mass of water in the solution = (150 - x)/1000 kg
Given molality = 0.15 = (x/342) / [(150 - x)/1000]
So 0.15 * (150 - x) = x * 1000/342
x = 7.319 gms
Moles of the sulphate = x/342 = 0.0214
Moles of Al⁺³ ions in the solution = 2 * 0.0214 = 0.0428
Total Number of ions = 0.0428 * 6.023 * 10²³ = 2.578 * 10²²
(multiplication by Avogadro number).
:-) :-) :-)
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