Question

If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain? Express your answer to two significant figures and include the appropriate units.

Answer 1

Explanation:

Let us assume that the given data is as follows.

        V = 3.10 L,        T = $19^{o}C$ = (19 + 273)K = 292 K

       P = 40 torr    (1 atm = 760 torr)

So,     P = $\frac{40 torr}{760 torr} \times 1 atm$

             = 0.053 atm

          n = ?

According to the ideal gas equation, PV = nRT.

Putting the given values into the above equation to calculate the value of n as follows.

                 PV = nRT

   $0.053 atm \times 3.10 L = n \times 0.0821 L atm/mol K \times 292 K$

                 0.1643 = $n \times 23.97$

                    n = $6.85 \times 10^{-3}$

It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.

               No. of moles = $\frac{mass}{\text{molar mass}}$

                 $6.85 \times 10^{-3} = \frac{mass}{46 g/mol}$  

                    mass = $315.1 \times 10^{-3}$ g

                              = 0.315 g

Thus, we can conclude that the mass of liquid ethanol is 0.315 g.