Amixture of helium and neon gases has a density of 0.1562 g/l at 40.2°c and 296 torr. what is the mole fraction of neon in this mixture?

You are given the density of helium and neon gas mixture which is 0.1562 grams per liters at 40.2degC and 296 torr. You are required to find the mole fraction of neon in this mmixture.

c = dRT/P

c = (0.1562g/L)(0.08206 L-atm/mol-K)(40.2+273)K/296torr(1atm/760torr)

c = 10.31 g/mol

since helium is 4 g/mol

mole fraction of neon = (10.31 - 4) / 10.31 = 0.612 

asking the goog might u get an answer

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