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3 years ago

9

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform ,

CHCl3, is 173.11 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 9.877 grams of the compound were dissolved in 150.9 grams of chloroform, the vapor pressure of the solution was 169.68 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound? chloroform = CHCl3 = 119.40 g/mol.

2 answers:

kogti [31]3 years ago

5 0

Answer:

387.3 g/mol.

Explanation:

Vapor Pressure of solution, Pₛₒₗ = X × Pₛₒₗᵥ

Where,

X = mole fraction of the solvent

Pₛₒₗᵥ = pressure of the solvent

Pₛₒₗᵥ = 173.11 mmHg

X = Pₛₒₗ/Pₛₒₗᵥ

= 169.68/173.11

= 0.9802.

Mole fraction is defined as the ratio of the number of moles of the substance to the total number of moles in the solution.

Mathematically,

X = Xₛₒₗᵥ/Xₛₒₗ

Number of moles of Chloroform = mass/molar mass

= 150.9/119.40

= 1.2638 mol

Therefore,

Total number of moles = 1.2638/0.9802

= 1.2893 mol

Number of moles of the compound = total number of moles - number of moles of Chloroform

= 1.2893 - 1.2638

= 0.0255 mol.

Molar mass of the compound = mass/number of moles

= 9.877/0.0255

= 387.3 g/mol.

alekssr [168]3 years ago

4 0

Answer:

The molecular weight of the unknown compound = 389.65 g/mol

Explanation:

Vapor Pressure of solution, P = mole fraction of solvent × Vapor Pressure of pure solvent = X × Pₛₒₗᵥ

P = 169.68 mmHg

Pₛₒₗᵥ = 173.11 mmHg

X = ?

X = P/Pₛₒₗᵥ = 169.68/173.11 = 0.9802

X = (number of moles of Chloroform)/(total number of moles)

Number of moles of Chloroform = mass/molar mass = 150.9/119.40 = 1.264 moles

X = 0.9802 = 1.264/(total number of moles)

Total number of moles = 1.264/0.9802 = 1.289 moles

Number of moles of unknown compound = total number of moles - number of moles of Chloroform = 1.289 - 1.264 = 0.0253 moles

Molar mass of unknown compound = (mass of unknown compound)/(number of moles of unknown compound) = 9.877/0.0253 = 389.65 g/mol

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$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

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