Question

A sample of damp air in a 1.00 L container exerts a total pressure of 741.0 torr at 20 oC; but when it is cooled to -10 oC, the water vapor freezes to ice with neglegible pressure giving a pressure of 607.1 torr. How many molecules of water were present in the gas

Answer 1

From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.

From the information we have;

Volume of the damp air =  1 L

Pressure of the damp air =  741.0 torr or 0.975 atm

Temperature of the gas = 20 oC + 273 = 293 K

R = 0.082 atm LK-1mol-1

Number of moles = ?

n =PV/RT

n = 0.975 × 1/0.082 ×  293

n = 0.041 moles

Volume of water vapor = 1 L

Temperature of water = -10 oC + 273 = 263 K

Pressure of the gas = 607.1 torr or 0.799 atm

R = 0.082 atm LK-1mol-1

n= PV/RT

n = 0.799 × 1/ 0.082 × 263

n = 0.037 moles

Number of moles of water = 0.041 moles -  0.037 moles = 0.004 moles

If 1 mole = 6.02 × 10^23 molecules

0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole

= 2.41 × 10^21 molecules

Learn more: brainly.com/question/2510654