Chlorodifluoromethane (CHF 2 Cl) was widely used in the compression/cooling circ

Question

Chlorodifluoromethane (CHF2Cl) was widely used in the compression/cooling circuits of refrigeration or air-conditioning systems. Since the discovery that such compounds (HCFCs and CFC's) released into the atmosphere were a major cause of depletion of stratospheric ozone, newer refrigeration systems make use of certain hydrofluorocarbons (HFCs), which are degraded in the lower atmosphere, instead. Often, mixtures of such compounds are used.

Suppose a sample of refrigerant gas consisting of a simple mixture of the gases pentafluoroethane (C2HF5) and 111-trifluoroethane (C2H3F3) has a density of 1.86 g/L at 33 °C and 0.432 atm.

Calculate the average molecular mass for this sample.

Calculate the volume percentage of C2H3F3 in the sample.

Explanation / Answer

Gas law equation can be used to determine the average molar mass of the sample.

From Gas law PV= nRT, where P= pressure in atm=0.432, n= no of moles= mass/molar mass, R= gas constant =0.0821 L.atm/mole. K, T= temperature in K= 33+273= 306K

Hence PV= (mass/molar mass)*RT

P* molar mass= (mass/volume)*RT= density*RT

Substituting the values given in the aboce equation

0.432*molar mass = 1.86g/L* 0.0821L.atm/mole.K* 306

Molar mass = 1.86*0.0821*306/0.432 g/mole=108.2 g/mole

Basis : 1 mole of mixture of C2HF5 and C2H3F3. For 1 mole = mass/molar mass, mass =molar mass

Let x= moles of C2HF5, 1-x =moles of C2H3F3

Atomic weights : C=12, H= 1 and F=19, molar masses (g/mole): C2H5F5=2*12+1+5*19=120

Molar mass of C2H3F3= 2*12+3*1+3*19=84

Molar mass of mixture = x*120+(1-x)*84= 108.2

120x+84- 84x= 108.2

36x= 108.2-84, x=0.67 since the temperature remains constant, the mole % = volume %

Hence volume % of C2HF5= 67% and that of C2H3F3= 100-67= 33%

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.