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avanturin [10]
2 years ago
15

Number of moles in 31.99g O2 And 176.8g NaCl

Chemistry
1 answer:
sergey [27]2 years ago
8 0

#O_2

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}

\\ \sf\longmapsto No\:of\:moles=\dfrac{31.99}{32}

\\ \sf\longmapsto No\:of\:moles\approx 1

#NaCl

\\ \sf\longmapsto No\:of\:moles=\dfrac{176.8}{58.5}

\\ \sf\longmapsto No\;of\:moles=3mol

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Answer:

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Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁)    (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

<h2>P₂ = 0.0166 mm Hg</h2>

Hope this helps

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Answer:

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