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1 year ago

A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atm at 20 ∘C . Calculate the van't Hoff factor i for AlCl3.

Chemistry

1 answer:

saw5 [17]1 year ago

4 0

Answer:

Van't Hoff factor, for this solution is 3.2

Explanation:

Osmotic pressure formula → π = M . R . T . i

π = Pressure → 3.85 atm

M = Molarity → 0.05mol/L

R = Constant for the Ideal Gases → 0.082 L.atm / mol.K

T = Absolute T° (K) → T°C + 273

20°C + 273 = 293K

Let's replace data:

3.85 atm = 0.05 mol/L . 0.082 L.atm/mol.K . 293K . i

3.85 atm / 0.05 . 0.082 atm . 293 = i

3.20 = i

If we assume complete ionization, i must be ≅ 4

AlCl₃ → Al³⁺ + 3Cl⁻

We have 1 mol of aluminum cation and 3 moles of chloride, 4 total moles.

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evablogger [386]

Answer:

Newton's Third Law

Explanation:

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1 year ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Mrrafil [7]

Answer :

The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.

<u>Percent Ionic Character :</u>

It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom

Given : Molecule is TiAl₃

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 ( From image shared )

Plug the value in above formula :

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

Value of e⁻¹ = 0.90

Percent ionic character = 1 - 0.90 * 100

Percent Ionic character = 10 %

<u>Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .</u>

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1 year ago

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galben [10]

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Explanation:

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1 year ago

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nataly862011 [7]

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1 year ago

For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

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2 years ago