The volume is 19.76987448 by taking the known variables mass=18.9g and density=0.956g/ml
To get volume you divide the mass by the density which gives you about
19.77 ml in volume
The reaction that should be followed is
Na2SO4 + C<span>a(NO3)2 --> CaSO4 + 2NaNO3</span>
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (<span>1.54×10−2 mol / L) = 1.155x10-3 mol
mol Ca(NO3)2 = 0.075 L (</span><span>1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2
so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 % </span>
They share a common ancestor, but theyre different species
Answer:
(D) Na₂SO₄•10H₂O (M = 286).
Explanation:
- The depression in freezing point of water by adding a solute is determined using the relation:
<em>ΔTf = i.Kf.m,</em>
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.
