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A liquid was known to be either methyl alcohol (CH3OH) or ethyl alcohol (CH3CH2OH). The Dumas method for determining molar masses was used to obtain an approximate molar mass, and this value was used to identify the alcohol. The gaseous alcohol at 98oC and 740Torr has a mass of 0.276g in a Dumas bulb of volume equal to 270mL. Which alcohol was present

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Answer:

Methyl alcohol was present.

Explanation:

To solve this problem we use the PV=nRT formula:

  • P = 740 Torr ⇒ 740/760 = 0.973 atm
  • V = 270 mL ⇒ 270/1000 = 0.270 L
  • T = 98 °C ⇒ 98 + 273.16 = 371.16 K

And solve for n:

  • n = PV/(RT)
  • n = 0.973 atm * 0.27 L / (0.082atm·L·mol⁻¹·K⁻¹ * 371.16 K)
  • n = 8.632 x 10⁻³ mol

Now we divide the mass and the moles to calculate the molar mass:

  • 0.276 g / 8.632 x 10⁻³ mol = 31.97 g/mol

Methyl alcohol (CH₃OH) has that molar mass.

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