10% 100

Assume that the volume of the mixture = 1liter, x is the number of mole of C2HF5 and y is the number of mol of the second one

The mass of the mixture[ m (mixture)] = V.d = 2.76*1=2.76 (g)

The number of mole of the mixture = pV/RT = 0.0378 (mol)

m(mixture) is the sum of the mass of 2 gases :210x + 52y = 2.76

n(mixture) = x + y = 0.0378

===>x =0.005 and y=0.0328

It is due to the fact that in a mixture of gases, %V = %n

===> %V of the first one = x*100/(x+y) = 0.005*100/0.0378

= 13.2%

%V of the second one = 100% - 13,2% = 86.8%

The mass of the mixture[ m (mixture)] = V.d = 2.76*1=2.76 (g)

The number of mole of the mixture = pV/RT = 0.0378 (mol)

m(mixture) is the sum of the mass of 2 gases :210x + 52y = 2.76

n(mixture) = x + y = 0.0378

===>x =0.005 and y=0.0328

It is due to the fact that in a mixture of gases, %V = %n

===> %V of the first one = x*100/(x+y) = 0.005*100/0.0378

= 13.2%

%V of the second one = 100% - 13,2% = 86.8%