Question

The following procedure provides a crude method of determining the molar mass of a volatile liquid. A liquid of mass 0.0224 g is introduced into a syringe and the end is capped (sealed). The syringe is transferred to a temperature bath maintained at 49.9 oC, and the liquid vaporizes. As the liquid vaporizes the plunger is pushed out. At equilibrium, the plunger reads 6.56 mL of gas. Atmospheric pressure is 740. mmHg. What is the approximate molar mass of the compound (in g/mol)?

Answer 1

Answer:

78.85 g/mol

Explanation:

The molecular weight can easily be calculated  from the ideal gas law if we have all the information since:

                  PV = nRT but   n= m/MW

therefore substituting and rearranging the equation we have

 PV= (m/MW) RT

MW= mRT/(PW)

attached  please find the calculation

Answer 2

Explanation:

We know that 740 mm Hg = 0.97368 atm

According to ideal behavior for the vapor,

                      PV = nRT

here,   R = 0.082 L $atm mol^{-1}K^{-1}$

            T = 273.15 + 62.7 = 335.85 K

            V = $6.81 \times 10^{-3}$ L

Therefore, first we will calculate the number of moles as follows.

             n = $\frac{PV}{RT}$

                = $\frac{0.97368 \times 6.56 \times 10^{-3}}{0.082 \times 322.9}$

                = $0.241 \times 10^{-3}$ moles

Now, molar mass of the given compound will be calculated as follows.

         Number of moles = $\frac{mass}{\text{molar mass}}$

         $0.241 \times 10^{-3} mol = \frac{0.0224 g}{\text{molar mass}}$

                 molar mass = 92.94 g/mol

Thus, we can conclude that molar mass of the compound is 92.94 g/mol.