First step: convert 22.4% into fraction
that is 22.4/100 which is equivalent to 0.224
The mass of Cucl2 contained in 75.85 of 22.4% by mass of solution of CUcl2 in water is therefore,
0.224 x75.85=16.99g
alternatively
75.85 --->100%
? 22.4%
by cross multiplication =(22.4 x 75.85)/100 =16.99g
C is correct. right now gravity holds the atmosphere so if there is no gravity then no atmospher
If it is a single cell then it is unicellular. it is an animal cell if it ha no cell wall and cells without nucleus are prokaryotes so it could be a Monera like bacteria.
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<h2 /><h2 /><h2><u>Solution </u><u>3</u><u> Is The Most Concentrated</u></h2>
<h3>S1:</h3>
M = m/v
= 100ml ÷ 2 spoons × 100%
= - 5,000 μg/ppb³
= <u>50% Diluted</u>
<h3>S2:</h3>
M = m/v
= 200ml ÷ 5 spoons × 100%
= - 4,000 μg/ppb³
= <u>40% Diluted</u>
<h3>S3:</h3>
M = m/v
= 300ml ÷ 6 spoons × 100%
= - 5,000 μg/ppb³
= <u>50% Diluted</u>
<h3>S4: </h3>
M = m/v
= 600ml ÷ 8 spoons × 100%
= - 75,000 μg/ppb³
= <u>75% Diluted</u>
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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.
