Question

How many grams ofAl2(SO4)3will be required to prepare 4 litres of 0.025M solution

Answer 1

Answer:

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

Explanation:

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).

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for now bye..........................