Question

2.00 g of an unknown volatile liquid was placed in a flask with a total volume of 265.0 mL. The cover of the flask has only a pinhole, which allows the internal pressure to equilibrate with the outside pressure. The temperature was raised to 100.00C by placing the flask in a constant temperature water bath. At this temperature, it was found that only 0.581 g of the liquid substance remained in the flask. A barometer in the lab showed the pressure to be 752 mm Hg. What's the molar mass of the unknown liquid?

Answer 1

Answer:

165.52 g/mol

Explanation:

Upon temperature was raised to 100 C part of the initial mass of liquid is gas, exactly: 2.00-0.581 = 1.419 g.  This gas occupy a volume of approximately 265 mL, its pressure of 752 mm Hg and its temperature is 100 C. Assuming that this gas behaves as gas ideal we can estimate the molecular weight $M_W$taking into account tha gas ideal law:

$PVM_{W}=RTm$  with R= 0.082 atm L/(mol K)

or $M_{W}=\frac{RTm}{PV}$

Before using of the formula, we need convert volume units to L (1L=1000 mL) pressure to atm (1 atm=760 mmHg) and temperature to K (K=273+C).

$265/1000 = 0.265 L$

$752/760=0.989 atm$

$273+100=373 K$

Finally, replacing in the formula

$\frac{0.082*1.419*373}{0.989*0.265}=165.52$ g/mol