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3 years ago

What is the value of freezing point depression of a solution of 15.0g of NaCl in 250g of water?

Chemistry

1 answer:

kobusy [5.1K]3 years ago

7 0

Answer:

-3.82ºC is the freezing point of solution

Explanation:

We work with the Freezing point depression to solve the problem

ΔT = m . Kf . i

ΔT = Freezing point of pure solvent - freezing point of solution

Let's find out m, molality (moles of solute in 1kg of solvent)

15 g / 58.45 g/mol = 0.257 moles of NaCl

NaCl(s) → Na⁺ (aq) + Cl⁻(aq)

i = 2 (Van't Hoff factor, numbers of ions dissolved)

m = mol /kg → 0.257 mol / 0.250kg = 1.03 m

Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)

0ºC - Tºf = 1.03m . 1.86ºC/m . 2

Tºf = -3.82ºC

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The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g)

wolverine [178]

Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478. The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59

A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.

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Answer:

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