Question

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

Answer 1

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 $C_{2}H_{5}OH$.