Agas mixture called heliox, 6.11% o2 and 93.89% he by mass, is used in scuba tanks for descents more than 65 m below the surface. calculate the mole fractions of he and o2 in this mixture. he is 0.9980

The mole fractions of He in the mixture is 0.992.

The mole fraction of in the mixture is 0.008.

Explanation:

A gas mixture called heliox, 6.11% and 93.89% He gas by mass by mass.

Percentage of oxygen gas in a mixture by mass = 6.11%

In 100 gram mixture 6.11% of oxygen is present by mass :

Percentage of helium gas in a mixture by mass = 93.89%

In 100 gram mixture 93.89% of helium is present by mass :

Mole fraction of component 'x' in a mixture of 'x' and 'y'=

= moles of gas 'x'=

= moles of gas 'y'=

Moles of oxygen gas ,when 6.11 grams of oxygen in present in a mixture:

Moles of helium gas ,when 93.89 grams of helium in present in a mixture:

The mole fractions of He in the mixture is 0.992.

The mole fraction of in the mixture is 0.008.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml