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Lana71 [14]
3 years ago
9

A 255 mL round--bonom flask is weighed and found to have a mass of 114.85 g. A few millimeters of an easily vaporized liquid are

added to the flask and the flask is immersed in a boiling water bath. All of the liquid vaporizes at the boiling tempentture of water, filling the flask with vapor. When all of the liquid has vaporized, the flask is removed from the bath, cooled, dried, and reweighed. The new mass of the flask and the condensed vapor is 115.23 g. Which of the following compounds could the liquid be?
a. C₄H₁₀b. C₃H₇OHc. C₂H₆d. C₂H₅OHe. C₄H₉OH
Chemistry
1 answer:
Alex17521 [72]3 years ago
4 0

Answer: Option (d) is the correct answer.

Explanation:

According to the given situation, mass of compound will be calculated as follows.

Mass of compound = mass of flask and condensed vapor - mass of flask

                                 = 115.23 - 114.85

                                = 0.38 g

Volume (V) = 255 mL = 255 \times 10^{-6} m^{3}        (as 1 ml = 10^{-6} m^{3})

Pressure (P) = 101325 Pa

Temperature = 100^{o}C = (100 + 273) K = 373 K

Now, according to the ideal gas equation, PV = nRT

and, moles of compound n = \frac{PV}{RT}

                     = \frac{101325 \times 255 \times 10^{-6}}{8.314 \times 373}

                     = 0.008332 mol

As, molar mass of compound = \frac{mass}{\text{no. of moles}}

                                                  = \frac{0.38}{0.008332}

                                                  = 46 g/mol

Therefore, the compound is C_{2}H_{5}OH (molar mass = 12 x 2 + 5 x 1 + 16 + 1 = 46 g/mol).

Thus, we can conclude that out of the given options the liquid could be C_{2}H_{5}OH.  

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Answer:

- Mole fraction of Chlorine Pentafluoride

= 0.265

- Partial Pressure of Chlorine Pentafluoride

= 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= 0.735

- Partial Pressure of Sulfur Hexafluoride

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Explanation:

First of, we calculate the number of moles of each gas present.

Number of moles = (Mass)/(Molar Mass)

For ClF₅

Mass = 4.28 g

Molar Mass = 130.445 g/mol

number of moles of Chlorine Pentafluoride

= (4.28/130.445) = 0.0328 moles

For SF₆

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Molar Mass = 146.06 g/mol

number of moles of Sulfur Hexafluoride

= (13.3/146.06) = 0.0911 moles

Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.

Using the ideal gas equation

PV = nRT

P = total pressure in the tank = ?

V = volume of the tank = 5.00 L = 0.005 m³

R = molar gas constant = 8.314 J/mol.K

T = temperature of the tank = 20.9°C = 294.05 K

n = total number of moles present = 0.1239 moles

P × 0.005 = (0.1239 × 8.314 × 294.05)

P = 60,580.45 Pa = 60.58 kPa.

- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)

- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)

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= (0.0328/0.1239) = 0.265

- Partial Pressure of Chlorine Pentafluoride

= 0.265 × 60.58 = 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= (0.0911/0.1239) = 0.735

- Partial Pressure of Sulfur Hexafluoride

= 0.735 × 60.58 = 44.53 kPa

Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa

Hope this Helps!!!

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And basic equilibrium of the conjugate base, is:

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