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andreev551 [17]
3 years ago
8

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point?(A) CuSO4•5H2O(M=250)(B)

NiSO4•6H2O(M=263)(C) MgSO4•7H2O(M=246)(D) Na2SO4 • 10 H2O (M = 286)
Chemistry
1 answer:
yarga [219]3 years ago
5 0

Answer:

  • <em>The solution of Na₂SO₄ . 10H₂O </em>( choice D)<em>, will have the lowest freezing point.</em>

Explanation:

1) The lowering of the freezing point is a colligative property which means that it depends, and can be calculated from some contants of the pure solvent, and the number of solute particles dissolved.

  • ΔTf = m × Kf × i

Where, ΔTf is the reduction in the freezing point, m is the molality of the solution, Kf is the cryoscopic constant of the solvent, and i is the Van't Hoff factor.

2) Find the molality of each solution, m:

  • Formulae:

       moles of solute, n = mass in grams / molar mass

       m = n / kg of solvent

(A) CuSO₄•5H₂O (M=250)

  • n = 20.0 g / 250 g/mol = 0.0800 mol

  • m = 0.0800 mol / 0.200 kg = 0.400 m

(B) NiSO₄•6H₂O(M=263)

  • n = 20.0 g / 263 g/mol = 0.0760 mol

  • m = 0.0760 mol / 0.200 kg = 0.380 m

(C) MgSO₄•7H₂O (M=246)

  • n = 20.0 g / 246 g/mol = 0.0813 mol

  • m = 0.0813 mol / 0.200 kg = 0.406 m

(D) Na₂SO₄ • 10 H₂O (M = 286)

  • n = 20.0 g / 286 g/mol = 0.0699 mol

  • m = 0.0699 mol / 0.200 kg = 0.350 m

3) Van't Hoff factor.

Since, all the solutes are ionic, you start assuming that they all dissociate 100%.

That means that:

  • Each unit of CuSO₄.5H₂O yields 2 ions in water ⇒ i = 2

  • Each unit of NiSO₄. 6H₂O yileds 2 ions in water ⇒ i = 2

  • Each unit of MgSO₄.7H₂O yields 2 ions in water ⇒ i = 2

  • Each unit of Na₂SO₄.10H₂O yields 3 ions in water ⇒ i = 3

4) Comparison

Being Kf a constant for the four solutions (same solvent), you just must compare the product m × i

  • CuSO₄.5H₂O: 2 × 0.400 = 0.800

  • NiSO₄. 6H₂O: 2 × 0.380 = 0.760

  • MgSO₄.7H₂O: 2 × 0.406 = 0.812

  • Na₂SO₄.10H₂O: 3 × 0.406 = 1.218

As you see from above calculations, the dissociation factor defines the situation, and you can conclude that the last choice, i.e. the solution of Na₂SO₄ . 10H₂O, will have the greatest decrease of the freezing point, resulting in the lowest freezing point.

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Explanation:

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Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0

Let energy change be E for 1 atom.

E=E_{\infty}-E_1=0-(-13.6  eV)=13.6 eV

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Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol

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Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-340eV)=340 eV

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Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 340eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol

E'=32,804.31 kJ/mol

The energy  required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ ,Li^{2+}atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0

Let energy change be E.

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1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol

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The energy  required to remove the electron in the ground state is 11,809.55 kJ/mol.

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Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV

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E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol

E'=820,107.88 kJ/mol

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