Asolution of 20.0 g of which hydrated salt dissolved in
200 g h2o will have the lowest freezing point?
(a) cuso4 • 5 h2o (m = 250)
(b) niso4 • 6 h2o (m = 263)
(c) mgso4 • 7 h2o (m = 246)
(d) na2so4 • 10 h2o (m = 286)

you so much for your and explanation : )

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, ΔTf is the depression in freezing point of water.

i is van't Hoff factor.

Kf is the molal depression constant.

m is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

Happy new years every one

Determine the  wavelength of light emitted when an electron  in an li2+ ion  makes a transition from an orbital with n  =  6 to an orbital with n  = 5.