let n1 and $$n_2bethemoleofCl_{35}andCl_{37}$ in mixture then
Average molar mass=35∗n1+37∗n2n1+n2
35.45=35∗n1+37∗n2n1+n2
n1n2=3.44
PV1=n1RT;PV2=n2RT
v1V2=n1n2
At constant P and T,
Volume of gas ∝ number of mole i.e.the mole ratio yields the volume ratio .
Therefore partition should be made in the volume ratio of 3.44: 1.Also, the pressure at this condition is same as the original condition since the volume of box and number of mole along with temperature are constant.
So, the value of x is 1.